9
9A 9B 9C 9D 9E 9F 9G 9H 9I 9J
Antidifferentiation Integration of ex, sin ((x) and cos ((x) Integration by recognition Approximating areas enclosed by functions The fundamental theorem of integral calculus Signed areas Further areas Areas between two curves Average value of a function Further applications of integration
Integration areaS of STudy
• Informal treatment of the fundamental theorem
• Antiderivatives of polynomial functions and of f (ax + b), where f is xn for n ∈ Q, ex, sin (x), cos (x) or linear combinations of these • Definition of the definite integral as the limiting b
a
n
∑ f ( xi* ) δ xi, δ x→0
value of a sum ∫ f ( x ) dx = lim
i =1
where the interval [a, b] is partitioned into n subintervals, with the ith subinterval of length δx δ i and containing xi*, and δx = max{δxi: i = 1, 2, … n} and evaluation of numerical approximations based on this definition • Informal approximation using areas under curves by left and right rectangles • Examples of the definite integral as a limiting value of a sum involving quantities such as area under a curve, distance travelled in a straight line and cumulative effects of growth such as inflation • Antidifferentiation by recognition that F ′( x ) = f ( x ) implies ∫ f ( x ) dx = F ( x ) + c
9a
b
of calculus, ∫ f ( x ) dx = F (b) − F (a) a
• Properties of antiderivatives and definite integrals: ( x ) ± bg( x )] dx = a ∫ f ( x ) ddxx ± b ∫ g( x ) dx ∫ [af (x b
∫a
b
∫a
a
∫a
c
c
f ( x ) dx +
∫b f ( x) dx = ∫a f ( x) dx
f ( x ) dx =
−
a
∫b
f ( x ) dx
f ( x ) dx = 0
• Application of integration to problems involving calculation of the area of a region under a curve and simple cases of areas between curves, such as distance travelled in a straight line; average value of a function; other situations modelled by the use of the definite integral as a limiting value of a sum over an interval; and finding a function from a known rate of change
antidifferentiation
eBoo k plus eBook Digital doc
10 Quick Questions
As we have seen, the process of differentiation enables us to find the gradient of a function. The reverse process, antidifferentiation (or integration), will find the function for a particular gradient. Integration has wider applications including calculation of areas, volumes, energy, probability and many more quantities in science and business. Note that d f ( x ) means differentiate f (x) with respect to x; that is, d f ( x ) = f ′( x ). dx dx
416
maths Quest 12 mathematical methods CaS for the Casio ClassPad
So f (x) is the antiderivative of f '(x), denoted as f ( x ) = ∫ f '( x ) dx where ∫ means antidifferentiate, or integrate, or find an indefinite integral and dx indicates that the integration of the function is with respect to x. Since
d (ax + c) = a, where a and c are constants dx
then
∫ a dx = ax + c
Since then
d ax n + 1 = ax n dx n + 1
∫ ax n dx =
ax n + 1 + c, n ≠ − 1 n +1
In the expression above, the term c is used to denote a constant. In the antiderivative of a function, there are an infinite number of possibilities for c. However, when we are finding an antiderivative, we set c to zero. That is, finding an antiderivative means ‘let c = 0’, or ‘do not add on the c’. For example, the antiderivative of 3x2 + 4x + 5 is x3 + 2x2 + 5x + c. An antiderivative of 2 3x + 4x + 5 is x3 + 2x2 + 5x.
Properties of integrals Since d is a linear operator, so too is ∫ ..Therefore, dx ∫ [ f ( x) ± g( x)] dx = ∫ f ( x) dx ± ∫ g( x) dx That is, each term can be integrated separately, and
∫ k f ( x) dx = k ∫ f ( x) dx That is, a ‘constant’ factor of the function can be taken to the front of the integral. So ∫ [af ( x) ± bg( x)]dx = a ∫ f ( x )dx ± b ∫ g( x )dx Worked Example 1
Antidifferentiate each of the following, expressing answers with positive indices. − 3 a 2x7 b 4x 3 c x Think a
b
c
Write
1
Integrate by rule; that is, add 1 to the index and divide by the new index.
2
Simplify.
1
Integrate by rule.
2
Simplify.
3
Express the answer with a positive index.
1
When a square root is involved, replace it with a fractional index.
a
∫ 2 x 7 dx =
2 x8 +c 8
x8 +c 4 − 4x 2 dx = − + c 2 =
b
∫ 4x
−3
−
c
∫
= −2x 2 + c − 2 = 2 x 3 3 dx = ∫ 1 dx x x2
Chapter 9 Integration
417
2
= 3∫ x
Bring the x to the numerator and change the sign of the index.
−1 2 dx
1
3x 2
+c
3
Integrate by rule.
=
4
Simplify.
= 6x 2 + c
5
Write the answer in the form it was given.
=6 x +c
1 2 1
Worked Example 2
Find the following indefinite integral. ∫(x. - 1)(3x + 5) dx Think
Write
∫ ( x − 1)(3x + 5) dx = ∫ (3x 2 − 3x + 5x − 5) dx = ∫ (3 x 2 + 2 x − 5) dx
1
Expand the expression.
2
Collect like terms.
3
Integrate each term separately.
=
4
Simplify each term.
= x3 + x 2 − 5x + c
3x3 2 x 2 + − 5x + c 3 2
Integration of (ax + b)n where n ≠ − 1 By applying the chain rule for differentiation: d (ax + b) n + 1 = a( n + 1)(ax + b) n dx
so
∫ a(n + 1)(ax + b)n dx = (ax + b)n + 1 + c
or
a( n + 1) ∫ (ax + b) n dx = (ax + b) n + 1 + c
or
∫ (ax + b)n dx =
(ax + b) n + 1 +c a( n + 1)
Worked Example 3 n+1
Antidifferentiate 4(5x − 2)3 by using ( ax + b) n dx = ( ax + b) ∫ a( n + 1) Think
418
+ c.
Write
∫ 4(5x − 2)3 dx = 4 ∫ (5x − 2)3 dx
1
Express as an integral and take 4 out as a factor.
2
Apply the rule where a = 5 and n = 3.
=
3
Simplify the antiderivative by cancelling the fraction.
= 5(5x − 2)4 + c
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
4(5 x − 2)4 +c 5(4) 1
Integration of x1 Since then
1 d log e ( x ) = dx x 1 ∫ x dx = loge ( x) + c, where x > 0
or
∫x
−1 dx
= log e x + c.
Worked Example 4
Antidifferentiate 4 . 7x Think 1
Write
4
4
= 2
1
∫ 7 x dx = ∫ 7 × x dx
Take 47 out as a factor.
4 7
1
∫ x dx
= 47 log e x + c
Integrate by rule.
- 1
Integration of (ax + b)
This can be done by applying the chain rule for differentiation: d a log e (ax + b) = , where a and b are constants. dx ax + b Multiplying both sides by 1 gives a 1 d a 1 × log e (ax + b) = × a dx ax + b a 1 ax + b 1 1 d log e ax + b dx dx = × ax + b a dx =
So
∫
so
∫ ( ax + b)
−1
dx =
1 log e ax + b + c a
1 Note that the a in the fraction a is the derivative of the linear function ax + b. Worked Example 5
Antidifferentiate
5 . 2x + 3
Think
Write
Method 1: Technology-free 1
Express as an integral and take 5 out as a factor.
2
Integrate by rule where a = 2.
5
1
∫ 2 x + 3 dx = 5 ∫ 2 x + 3 dx =
5 log e 2 x + 3 + c 2
Chapter 9 Integration
419
Method 2: Technology-enabled 1
On the Main page, using the soft keyboard, tap: • ) • • P Complete the entry line as: 5 ∫ ( 2 x + 3 ) dx Press E.
2
Write the answer and put in the constant of integration, as the CAS calculator omits it. Also note that the CAS uses ln instead of loge.
5 5 ∴ ∫ dx = log e 2 x + 3 + c 2 2 x + 3
Worked Example 6
Find ∫
6x + 5 dx . x2
Think
Write
∫
6x + 5 6x 5 dx = ∫ 2 + 2 dx 2 x x x
1
Express as separate fractions.
2
Simplify each fraction.
= ∫ (6 x
3
Integrate each term separately by rule.
= 6 log e x +
4
Simplify leaving the answer with positive indices.
= 6 log e x − 5 x + c 5 = 6 log e x − + c x
−1
−
+ 5 x 2 ) dx 5x
−1
−1 −1
+c
Worked Example 7
Find the equation of the curve g(x) given that g′(x) = 3 x + 2 and the curve passes through (1, 2). Think
420
Write
1
Write the rule for g′(x).
g′(x) = 3 x + 2
2
Rewrite g′(x) in index form.
g′(x) = 3 x 2 + 2
3
Express g(x) in integral notation.
g(x) =
4
Antidifferentiate to obtain a general rule for g(x).
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
1
1
∫ 3x 2 + 2 dx 3
= 3x 2 ÷ 3 + 2 x + c 2
5
Simplify.
3 2
= 3x × 2 + 2 x + c 3 1
3
g(x) = 2 x 2 + 2 x + c
6
Substitute coordinates of the given point into g(x).
7
Find the constant of antidifferentiation, c.
8
State the rule for g(x) in the form that it is given.
3
As g(1) = 2, 2(1) 2 + 2(1) + c = 2 2+2+c=2 so c = −2 3
g(x) = 2 x 2 + 2 x − 2
= 2 x3 + 2x − 2
Worked Example 8
If a curve has a stationary point (2, 3), and a gradient of 2x − k, where k is a constant, find: a the value of k b y when x = 1. Think a
1
2
Write
The gradient is dy so write the rule for the dx gradient.
a
Let dy = 0 (as stationary points occur when dx the derivative is zero) and substitute the value of x into this equation.
dy = 2x − k dx For stationary points, dy = 0, so 2x − k = 0 dx 2(2) − k = 0 as x = 2
b
3
Solve for k.
1
Rewrite the rule for the gradient function, using the value of k found in a above.
2
Integrate to obtain the general rule for y.
4 − k = 0 so k = 4 b
dy = 2x − 4 dx y = ∫ (2 x − 4)dx = x2 − 4x + c
3
Substitute the coordinates of the given point on the curve to find the value of c.
Since curve passes through (2, 3), 3 = 22 − 4(2) + c 3=4−8+c c=7
4
State the rule for y.
So y = x2 − 4x + 7
5
Substitute the given value of x and calculate y.
When x = 1, y = (1)2 − 4(1) + 7 =4
The relationship between the graph of an antiderivative function and the graph of the original function f (x) is the antiderivative of f ′(x) and is written as f ( x ) = ∫ f ′( x )dx. f ′(x) is the gradient function of the antiderivative function f (x).
Chapter 9 Integration
421
The graph of the antiderivative function f (x) can be derived from the graph of f ′(x) and results in a family of curves. For example if f ′(x) = 1 then the antiderivative function is f (x) = x + c, where c can take any real value.
Sketching the antiderivative function from the graph of the original function 1. The general shape of the graph of the antiderivative function can be determined from the graph of a polynomial function by increasing the degree by one. For example, if f ′(x) is a quadratic function, then f (x) is a cubic function. 2. The x-intercepts of f ′(x) become the turning points on the graph of f (x). 3. When f ′(x) is above the x-axis, the gradient of f (x) is positive. 4. When f ′(x) is below the x-axis, the gradient of f (x) is negative. Worked Example 9
Sketch the graph of the antiderivative function from the graph of the gradient function f ′(x) shown. f ′(x)
(1, 0)
(−2, 0) 0
x
(0, −1) Think
Write
1
State the shape of the antiderivative function.
The antiderivative function will be a positive cubic function.
2
Find the x-intercepts of the gradient function, f ′(x), and hence find the x-coordinates of the turning points.
There are x-intercepts when x = −2 and when x = 1, so f (x) has turning points when x = −2 and x = 1.
3
Find when the given graph, f ′(x), is above the x-axis and hence find when f (x) has a positive gradient.
f ′(x) is above the x-axis when x < −2 and when x > 1, so f (x) has a positive gradient when x < −2 and when x > 1.
4
Find when the given graph, f ′(x), is below the x-axis and hence find when f (x) has a negative gradient.
f ′(x) is below the x-axis when −2 < x < 1, so f (x) has a negative gradient when −2 < x < 1.
5
Sketch the curve.
The graph could be any of the family of graphs formed by vertical translations of the graph shown. f(x)
−2
422
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
0
1
x
REMEMBER
1.
d f ( x ) = f ′( x ) dx
2. f ( x ) = ∫ f ′( x ) dx 3. ∫ a dx = ax + c 4. ∫ ax n dx =
ax n + 1 + c, n ≠ − 1 n +1
5. ∫ [ f ( x ) ± g( x )] dx =
∫ f ( x) dx ± ∫ g( x) dx
6. ∫ k f ( x ) dx = k ∫ f ( x ) dx 7. ∫ (ax + b) n dx =
(ax + b) n + 1 + c, n ≠ − 1 a( n + 1)
1 dx = log e x + c, x ≠ 0 or ∫ x −1 dx = log e x + c x
8. ∫
1 1 1 dx = log e ax + b + c or ∫ (ax + b)−1 dx = log e ax + b + c ax + b a a 10. Write the answer in the same form as the question unless otherwise stated. 11. To draw the graph of the antiderivative function f (x) from the graph of the gradient function f ′(x), use the x-intercepts as the x-coordinates of the turning points, and make the gradient positive when f ′(x) is above the x-axis and negative when f ′(x) is below the x-axis. 12. For a polynomial function, the graph of f (x) is one degree greater than the graph of f ′(x). 9. ∫
Exercise
9a
Antidifferentiation 1 WE1 a x
Antidifferentiate each of the following, giving answers with positive indices. c x7 d 3x5 b x4
−
e 5x 2 x4 i 5
f
x3 j 2
2
3
m x 3 q
n 4 x 4
9 x2
2 WE2
−
−2x4
r
− 10
x6
g −6x 4 − x 4 k 3
h 2 x
o x
p
s
l
−3 7
8 x
t
x 5 x3 −6
(x x )
Find the following indefinite integrals.
a ∫ (2 x + 5) dx
b ∫ (3 x 2 + 4 x − 10) dx
c
∫ (10 x 4 + 6 x3 + 2) dx
d ∫ ( − 4 x 5 + x 3 − 6 x 2 + 2 x ) dx
e
∫ ( x3 + 12 − x 2 ) dx
f
∫ ( x + 3)( x − 7) dx
g ∫ 5( x 2 + 2 x − 1) dx
h ∫ ( x 2 + 4)( x − 7) dx
i
∫ x( x − 1)( x + 4) dx Chapter 9 Integration
423
3 MC
∫ ( x 2 + x + 2) dx is equal to:
A ∫ x 2 dx + x + 2
B ∫ x 2 dx + ∫ x dx + ∫ 2 dx
C ∫ (x 2 + x ) dx + 2
D x 2 + ∫ ( x + 2) dx
E x 2 + x + ∫ 2 dx 4 MC
∫ x( x + 3) dx is equal to:
A ∫ x dx ∫ ( x + 3) dx
B x ∫ ( x + 3) dx
C ( x + 1) ∫ x dx
D ∫ x dx + ∫ ( x + 3) dx
E
∫ ( x 2 + 3x) dx
5 WE3 Antidifferentiate each of the following by using ∫ (ax + b) n dx = a (x + 3)2 c 2(2x + 1)4 e (6x + 5)4 g (4 − x)3 i 4(8 − 3x)4 − k (2x + 3) 2 − m 6(4x − 7) 4 − o (6 − 5x) 3 6 MC
b d f h j l n p
(ax + b) n + 1 + c. a( n + 1)
(x − 5)3 −2(3x − 4)5 3(4x − 1)2 (7 − x)4 −3(8 − 9x)10 − (6x + 5) 3 − (3x − 8) 6 −10(7 − 5x)−4
∫ 3( x + 2)4 dx is equal to:
A 3 + ∫ ( x + 2)4 dx
B ∫ 3 dx + ∫ ( x + 2)4 dx
C 3 ∫ ( x + 2)4 dx
D 3 ∫ dx × ∫ ( x + 2)4 dx
E ( x + 2)4 ∫ 3 dx 7 WE4, 5 Antidifferentiate the following.
424
a
∫ x dx
3
b
∫ x dx
8
c
∫ 5x dx
d
∫ 3x dx
7
e
∫ 7x dx
4
f
∫ x + 3 dx
g
∫ x + 3 dx
3
h
∫ x + 4 dx
i
∫ x + 5 dx
j
∫ 3x + 2 dx
4
k
∫ 5x + 6 dx
8
l
∫ 2 x − 5 dx
o
∫ 5 − x dx
−5
m
∫ 3 + 2x
p
3 ∫ 6 − 11x dx
dx
n q
−2
−2
∫ 6 + 7x dx −2
∫ 4 − 3x
dx
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
r
6
1
−6
3
1
−8
∫ 5 − 2x dx
6
∫ x + 5 dx is equal to:
8 mC A 6∫
1 dx x+5
D
∫ 6dx
B E
∫ ( x + 5) dx
9 We6 Find ∫
1
∫ 6dx ∫ x + 5 dx
∫ 3(4 x + 1)
j
∫
m eBook plus Digital doc
SkillSHEET 9.1 Substitution and evaluation
∫
x+
∫ 6dx + ∫ x + 5 dx
c
∫
f
∫ 4(2 x − 5)5 dx
i
∫
( x − 5)( x + 3) dx x3
l
∫
x2 + x4 dx x
6
∫ ( x + 5) dx
(2 x + 7) dx. x
10 For the following mixed sets, find: 1 a ∫ x4 + 2x + b ∫ (3 x + 1)5 dx x −5 3 d ∫ e ∫ dx dx 2x + 1 6 − 10x g
1
C
−3
( x + 4) 2 dx 2x −1 3 k ∫ 5 x 2 − 2 x + 3 x 3 dx
h
dx
2 dx 3 − x
x2 + 2x − 1 x
n
dx
∫
3x 2 + 2 x − 1 dx x2
10 − x + 2 x 4 dx ∫ x3
11 We7 Find the equation of the curve f (x) given that: a f ′(x) = 4x + 1 and the curve passes through (0, 2) b f ′(x) = 5 − 2x and the curve passes through (1, −1) − c f ′(x) = x 2 + 3 and the curve passes through (1, 4) d f ′(x) = x + x and f (4) = 10 1
e f ′(x) = x 3 − 3 x 2 + 50 and f (8) = −100 1 f f ′(x) = − 2 x and f (1) = −5 x g f ′(x) = (x + 4)3 and the curve passes through (−2, 5) − h f ′(x) = 8(1 − 2x) 5 and f (1) = 3 −1 i f ′(x) = (x + 5) and the curve passes through (−4, 2) 8 j f ′′(x) and f (3) = 7 ( x) = 7 − 2x 12 We8 If a curve has a stationary point (1, 5), and a gradient of 8x + k, where k is a constant, find: a the value of k b y when x = −2. 13 A curve g(x) has g ′( x ) = kx +2 x , where k is a constant, and a stationary point (1, 2). Find: x a the value of k b g(x) c g(4). 14 We9 Sketch the graph of the antiderivative functions from each of the following graphs. a b c f ′(x) f′(x) f ′(x)
(−1, 0)
(1, 0) x
0
(−2, 0) 0
x
0
(0, 2)
(0, −3)
x
(0, −1)
Chapter 9
Integration
425
f ′(x)
d
e
f
f ′(x)
f ′(x)
(0, 1) (0, 0)
( 1–2 , 0) 0 (1, 0)
0
x
0
x
x
(0, −1)
g
f ′(x)
(−2, 0) (0, 0)
9B
(1, 0)
x
Integration of e x, sin (x) and cos (x) Integration of the exponential function e x Since
d x (e ) = e x dx
then
∫ e x ddxx = e x + c
and or
Therefore,
d kx (e ) = kke kx , where k is a constant dx
∫ kekx dx = ekx + c k ∫ e kx dx = e kx + c 1 ∫ ekx ddx = k ekx + c.
Worked examPle 10
Antidifferentiate each of the following. − e 5x 4x a 3e b c (e x − 1)2 4 ThInk a
b
426
WrITe
1
Integrate by rule where k = 4.
2
Simplify.
1
Rewrite the function to be integrated so that the coefficient of the e term is clear.
2
Integrate by rule where k = −5.
a
b
3e 4 x +c 4 3 = 4 e4 x + c
∫ 3e4 x ddxx = ∫
maths Quest 12 mathematical methods CaS for the Casio ClassPad
e
−5 x
4
dx = =
1
−
∫ 4 e 5 x ddx 1 4
e −
−5 x
5
+c
3
1 = 4e
Simplify the antiderivative.
−5x 5x
1 = − 20 e
c
1
Expand the function to be integrated.
2
Integrate each term by the rule.
c
×
−5 x
1 −5
+c
+c
∫ (e x − 1)2 dx = ∫ (e2 x − 2e x + 1) dx = 12 e 2 x − 2e x + x + c
Integration of trigonometric functions Since
d [sin (ax)] = a cos (ax) dx
and
d [cos (ax )] = − a sin (ax ) dx
it follows that
∫ sin ( ax) dx =
−
1 cos ( ax ) + c a
1
∫ cos ( ax) dx = a sin ( ax) + c Worked examPle 11
Antidifferentiate the following. a sin (6x)
b 8 cos (4x)
c 3 sin
− x 2
ThInk
WrITe
a
Integrate by rule.
a
b
1
Integrate by rule.
b
2
Simplify the result.
1
Integrate by rule.
2
Simplify the result.
c
−
1
∫ sin (6 x)dx = 6 cos (6 x) + c 8 ∫ 8 cos (4 x)dx = 4 sin (4 x) + c = 2 sin (4x) + c
c
− x dx = 2
∫ 3 sin
−
3 1 2
−
− x cos + c 2
− x = 6 cos + c 2
Worked examPle 12
Find ∫ [ 2 e 4 x − 5 sin ( 2 x ) + 4 x ] dx. ThInk
WrITe
Method 1: Technology-free 1
Integrate each term separately.
∫ [2e4 x − 5 sin (2 x) + 4 x]dx = 42 e 4 x −
2
Simplify each term where appropriate.
−5 2
cos (2 x ) + 42 x 2 + c
= 12 e 4 x + 25 cos (2 x ) + 2 x 2 + c
Chapter 9
Integration
427
Method 2: Technology-enabled 1
On the Main page, using the soft keyboard, tap: • ) • • P Complete the entry line as:
∫ 2e4 x − 5 sin (2 x) + 4 x dx
Press E. Note: When integrating trigonometric functions, ensure the CAS is set to Radian mode.
2
∴∫ (2e 4 x − 5 sin (2 x ) + 4 x ) dx
Write the answer and put in the constant of integration.
e 4 x 5 cos (2 x ) + + 2x2 + c 2 2
=
REMEMBER
Exercise
9b
1
1.
∫ e x dx = e x + c and ∫ ekx dx = k ekx + c
2.
∫ sin (ax) dx =
−1
a
cos (ax ) + c and ∫ cos (ax ) dx =
1 sin (ax ) + c a
Integration of e x, sin (x) and cos (x) 1 WE10 a e2x − d e 3x e6 x g 2 j
−8e−2x x
m 3e 2 e x − e− x p 2
Antidifferentiate each of the following. b e4x e 5e5x 2e 3 x h 3 x
−x 3
i
−3e6x x
l 0.1e 4
k e 3 n 3e
c e−x f 7e4x
o e x + e−x
2 Find an antiderivative of (1 + ex)2. 3 Find an antiderivative of (ex − 1)3. 4 Find an antiderivative of x3 − 3x2 + 6e3x. 5 MC If f ′(x) = e2x + k and f (x) has a stationary point (0, 2), where k is a constant, then: a k is equal to: A e B e2 C 1 D −1 E −e b f (1) is equal to: A e2 − 1 B e 2 + 12 C 12 e 2 + 12 D e4 E 1 12
428
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
6 WE11 Antidifferentiate the following. a sin (3x) b sin (4x) cos ( 2 x ) d e sin (−2x) 3 4 sin (6 x ) g h 8 cos (4x) 3 x j −2 cos (−x) k sin 3 − x x m 3 sin n − 2 sin 5 4 − x p − 6 cos 2 s
− 2 sin
5x 2
t
− 3 cos
−6
sin (3x) x l cos 2 x o 4 cos 4
3x r 6 cos 4
7x 4
u 5 sin (πx) − 4x s x − sin π
b [sin (2x ) − cos ( x )] dx ∫
∫ [cos (4 x) + sin (2 x)] dx e ∫ [4 cos (4 x ) − 13 sin (2 x)] dx c
i
π x w − 2 cos 3
7 WE12 Find: a [sin ( x ) + cos ( x )] dx ∫
∫ 3 sin ( π2x
f cos (−3x)
2x q 4 sin 3
π x v 3 cos 2
g
c cos (7x)
) + 2 cos ( π3x ) dx
) − cos (2x) dx
d
∫ sin ( 2x
f
∫ [5x + 2 sin ( x)] dx
h ∫ [3e6 x − 4 sin (8 x ) + 7] dx
8 Find the antiderivative of e4x + sin (2x) + x3. 9 Find an antiderivative of 3x2 − 2 cos (2x) + 6e3x. 10 Antidifferentiate each of the following. 1 a x 3 − b x2 + 4 cos (2x) − e−x + e2 x 2x + 3 x x c sin + e 2 − (3 x − 1)4 3
d
x x x e 3 sin − 2 cos − e 5 3 2 −
f
1 x + e 4 x + cos 5 3x − 2 π x x + 2 x − 2 sin + 5 3
11 In each of the following, find f (x) if: π a f ′(x) = cos (x) and f = 5 b f ′(x) = 4 sin (2x) and f (0) = −1 2 x x x c f ′( x ) = 3 cos and f (π ) = 9 2 d f ′( x ) = cos − sin and f (2π) = −2. 2 4 4 dy π x 12 If = sin + k, where k is a constant, and y has a stationary point (3, 4), find: 6 dx a the value of k b the equation of the curve c y when x = 6. 13 A curve has a gradient function f ′(x) = 4 cos (2x) + ke x, where k is a constant, and a stationary point (0, −1). Find: a the value of k b the equation of the curve f (x) π c f correct to 2 decimal places. 6
Chapter 9 Integration
429
9C
Integration by recognition As we have seen, if d [ f ( x )] = g( x ) dx then
∫ g( x) dx = f ( x) + c, where g( x) = f '( x).
This result can be used to determine integrals of functions that are too difficult to antidifferentiate, via differentiation of a related function.
Worked examPle 13
eBook plus
a Find the derivative of the function y = (5x + 1)3. b Use this result to deduce the antiderivative of 3(5x + 1)2.
Tutorial
int-0564
ThInk a
b
WrITe
1
Write the function and recognise that the chain rule can be used.
2
Let u equal the function inside the brackets. du Find . dx
3 4
Express y in terms of u.
5
Find
6
Write the chain rule.
7
Find
8
Replace u with the expression inside the brackets and simplify where applicable.
1
2
3
Worked example 13
a y = (5x + 1)3
Let u = 5x + 1 du =5 dx y = u3
dy . du
dy = 3u 2 du dy dy du = × dx du dx dy = 3u 2 × 5 dx
dy using the chain rule. dx
du dx = y + c1, express the relationship dx in integral notation. dy Remove a factor from so that it resembles dx the integral required. Since ∫
Divide both sides by the factor in order to obtain the integral required.
= 15(5x + 1)2 b
∫ 15(5x + 1)2 ddxx = (5x + 1)3 + c1 5 ∫ 3(5 x + 1)2 dx = (5 x + 1)3 + c1
∫ 3(5x + 1)2 dx = 15 (5x + 1)3 + c, where c =
c1 5
Therefore, the antiderivative of 3(5x + 1)2 is 15 (5x ( 5x ++ 11))33 + cc..
Note that the shorter form of the chain rule below can be used to differentiate. If f ( x ) = [ g( x ))]]n then f ′ ( x ) = ng ′ ( x )[ g( x ))]]n − 1 .
430
maths Quest 12 mathematical methods CaS for the Casio ClassPad
Worked examPle 14 a Differentiate e x . 3
b Hence, antidifferentiate 6x2 e x . 3
ThInk a
1
Write the equation and apply the chain rule to differentiate y.
2
Let u equal the index of e.
3
Find
4
Express y in terms of u.
5
Find
6
b
WrITe a
y = ex
3
Let u = x3
du . dx
du = 3x 2 dx y = eu
dy . du dy Find using the chain rule and dx replace u.
dy = eu du dy = 3 x 2 eu dx = 3x 2 e x
dy in integral notation. dx
b
1
Express
2
Multiply both sides by a constant to obtain the integral required.
3
∫ 3x 2 e x dx = e x 3
3
+ c1
2∫ 3 x 2 e x dx = 2e 2e x + 2c1 3
3
2e x ∫ 6 x 2 e x dx = 2e 3
3
+ c, c where c = 2c1.
Therefore, the antiderivative of 6 x 2 e x is 2e x + c. c 3
3
Note that the shorter form of the chain rule below can be used to differentiate. If y = e f ( x ) then
dy = f ′( x)e f ( x ) . dx
Worked examPle 15 a Find the derivative of sin (2 x + 1) and use this result to deduce the antiderivative of 8 cos (2 x + 1). b Differentiate loge(5x2 − 2) and hence antidifferentiate ThInk a
b
x . 5 x2 − 2 WrITe a f (x) = sin (2x + 1)
1
Define f (x).
2
Differentiate using f ′(x) = g′(x) cos [g(x)] where f (x) = sin [g(x)].
3
Express f (x) using integral notation.
4
Multiply both sides by whatever is necessary for it to resemble the integral required.
4 ∫ 2 cos (2 x + 1) 1)dx = 4 sin (2 x + 1) + c
5
Write the integral in the form in which the question is asked.
∫ 8 cos (2 x + 1)dx = 4 ssinin (2 x + 1) + c
1
Define f (x).
f ′(x) = 2 cos (2x + 1)
∫ 2 cos (2 x + 1)dx = sin (2 x + 1) + c1
The antiderivative of 8 cos (2x + 1) is 4 sin (2x + 1) + c. b f (x) = loge (5x2 − 2)
Chapter 9
Integration
431
2
Differentiate using f ′( x ) = where f (x) = loge [g(x)].
g ′( x ) g( x )
3
Express f (x) using integral notation.
4
Take out a factor so that f ′(x) resembles the integral required.
5
Divide both sides by the factor to obtain the required integral.
10 x 5x 2 − 2
f ′( x ) =
10 10x
∫ 5x 2 − 2 dx = loge 10 ∫
5 x 2 − 2 + c1
x dx = log e 5 x 2 − 2 + c1 −2
5x 2
x
∫ 5x 2 − 2 dx = 101 loge
5x 2 − 2 + c
The antiderivative of x is 1 log e 5 x 2 − 2 + c. 5 x 2 − 2 10
Worked examPle 16
Differentiate x cos (x) and hence find an antiderivative of x sin (x). ThInk
WrITe
Let y = x cos (x)
1
Write the rule.
2
Apply the product rule to differentiate x cos (x).
3
Express the result in integral notation. (Do not add c, as an antiderivative is required.)
∴ ∫ [cos ( x ) − x sin sin ( x )] ddxx = x cos ( x )
4
Express the integral as two separate integrals.
5
Simplify by integrating. (Do not add c.)
6
Make the expression to be integrated the subject of the equation.
cos ( x ) dx − ∫ x ssin in ( x ) ddxx = x cos ( x ) ∫ cos sin ( x ) − ∫ x ssin in ( x ) ddxx = x cos ( x ) − x sin ( x ) dx = x cos ( x ) − sin ( x ) ∫
7
Simplify.
dy = x [−sin (x)] + [cos (x)](1) dx = −x sin (x) + cos (x) = cos (x) − x sin (x)
x ) dx = ssin in ( x ) − x cos ( x ) ∫ x sin ( x) Therefore, an antiderivative of x sin (x) is sin (x) − x cos (x).
Worked examPle 17 a Show that
5x + 1 4 = 5− . x+1 x+1
b Hence, find
ThInk a
432
Use algebraic long division to divide the numerator into the denominator.
∫
5x + 1 dx. x+1 WrITe
)
5
a x + 1 5x + 1
5x + 5 −4
maths Quest 12 mathematical methods CaS for the Casio ClassPad
So
5x + 1 4 = 5− x +1 x +1
b
b
∫
5x + 1 4 dx = ∫ 5 − x +1 x + 1
1
Write the expression using integral notation.
2
Express as two separate integrals.
= ∫ 5 dx − ∫
3
Antidifferentiate each part.
= 5 x − 4 log e x + 1 + c
4 dx x +1
rememBer
1. To differentiate using the chain rule, use one of the following rules. (a) If f (x) = [g(x)]n then f ′(x) = ng′(x)[g(x)]n − 1 (b) If y = e f (x),
dy = f ′( x )e f ( x ) dx
dy dy du = × dx du dx (d) f ′(x) = g′(x) cos [g(x)] where f (x) = sin [g(x)] f ′(x) = −g′(x) sin [g(x)] where f (x) = cos [g(x)] (c)
(e) f ′( x ) =
g ′( x) where f ( x ) = logg e g( x ) . g( x )
2. To antidifferentiate, use ∫ g( x ) dx = f ( x ) + c where g(x) = f ′(x).
exerCISe
9C
Integration by recognition 1 We13 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i (3x − 2)8 ii 12(3x − 2)7 2 5 b i (x + 1) ii 5x(x2 + 1)4 1 c i 2x − 5 ii 2x − 5 d i
4x + 3
+ 3x − 1 f i 2 x −1 e i
(x2
ii 7)4
2 4x + 3
ii (2x + 3)(x2 + 3x − 7)3 4x ii 2 ( x − 1)2
2 mC The derivative of (x + 7)4 is 4(x + 7)3. a Therefore, the antiderivative of 4(x + 7)3 is: A (x + 7)4 + c B 14 (x + 7)4 + c D 3(x + 7)4 + c E 12(x + 7)4 + c 3 b The antiderivative of (x + 7) is: A (x + 7)4 + c B 14 (x + 7)4 + c D 3(x + 7)4 + c
C 4(x + 7)4 + c
C 4(x + 7)4 + c
E 12(x + 7)4 + c
Chapter 9
Integration
433
3 mC If the derivative of (2x − 3)6 is 12(2x − 3)5, then ∫ 6(2 x − 3)5 dx is: A 2(2x − 3)6 + c B 4(2x − 3)6 + c C (2x − 3)6 + c 1 6 6 D 6(2x − 3) + c E 2 (2x − 3) + c 4 We14
For each of the following differentiate i and hence antidifferentiate ii.
e4x − 5
a i c i ex
ii 2e4x − 5 ii x e x
2
b i e6 − 5x d i ex − x
2
ii 10e6 − 5x ii (1 − 2 x )e x − x
2
2
5 We15 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i sin (x − 5) ii cos (x − 5) b i sin (3x + 2) ii 6 cos (3x + 2) c i cos (4x − 7) ii sin (4x − 7) d i cos (6x − 3) ii 3 sin (6x − 3) e i sin (2 − 5x) ii 10 cos (2 − 5x) f i cos (3 − 4x) ii −2 sin (3 − 4x) 12 x 20 g i loge (5x + 2) ii h i loge (x2 + 3) ii 2 5x + 2 x +3 i i loge (x2 − 4x) ii
x−2 x2 − 4x
6 We16
Differentiate i and hence find an antiderivative of ii. 2[ x cos cos ( x ) − sin ( x )] sin ( x ) a i x cos (x) + 2 sin (x) ii x sin (x) b i ii x x2 c i e x sin (x) ii 3e x [sin (x) + cos (x)] d i x sin (x) ii x cos (x) e i x ex ii x ex
7 For each of the following differentiate i and use this result to antidifferentiate ii. a i (2x − 3x2)6 ii 6x5(1 − 3x)(2 − 3x)5 x3 + 2x
b i 8 We17
a Show that
9 a Show that 10
x3 + 2x
3x − 2 1 =3+ . x −1 x −1
8x − 7 5 =4+ . 2x − 3 2x − 3
6x − 5 − 4 = 3+ . 3 − 2x 3 − 2x 2
b Hence, find ∫
3x − 2 dx. x −1
b Hence, find ∫
5x + 8 dx. x+2
b Hence, find ∫
8x − 7 dx. 2x − 3
b Hence, find ∫
6x − 5 dx. 3 − 2x
If y = loge[cos (x)]: a find
dy . dx
13 Differentiate 14
3x 2 + 2
5x + 8 2 =5− . x+2 x+2
a Show that
11 a Show that 12
ii
b Hence, find ∫ ta tan ( x ) dx. cos ( x ) 1 and hence find an antiderivative of 2 . sin ( x ) sinn ( x )
Differentiate loge (3x2 − 4) and hence find an antiderivative of
x . −4
3x 2
15 Differentiate sin (ax + b) and hence find an antiderivative of cos (ax + b). (Here, a and b are constants.) 16 Differentiate cos (ax + b) and hence find an antiderivative of sin (ax + b). (Here, a and b are constants.)
434
maths Quest 12 mathematical methods CaS for the Casio ClassPad
17 Differentiate eax + b and hence find an antiderivative of eax + b. (Here, a and b are constants.) 18 Antidifferentiate each of the following. a sin (3π x + 1) b cos (1 − 4π x) π x πx d sinn 2 + e 3 cos + 5 2 3
9d
eBook plus
c eπ x + 3
Digital doc
f cos (x)esin (x)
WorkSHEET 9.1
approximating areas enclosed by functions
eBook plus Interactivity
int-0254 Approximating areas enclosed by functions
There are several ways of finding an approximation to the area between a graph and the x-axis. We will look at two methods: 1. the left rectangle method 2. the right rectangle method
The left rectangle method Consider the area between the curve f (x) shown at right, the x-axis y f(x) and the lines x = 1 and x = 5. If the area is approximated by ‘left’ rectangles of width 1 unit then the top left corner of each rectangle touches the curve at one point. So, the height of rectangle R1 is f (1) units and the area of R1 = 1 × f (1) square units (area of a R1 R2 R3 R4 rectangle = height × width). Similarly, the area of R2 = 1 × f (2) square units, 0 1 2 3 4 5 x the area of R3 = 1 × f (3) square units, the area of R4 = 1 × f (4) square units. Therefore, the approximate area under the graph between the curve f (x), the x-axis and the lines x = 1 to x = 5 is 1[ f (1) + f (2) + f (3) + f (4)] square units, (the sum of the area of the four rectangles). If the same area was approximated using rectangle widths of 0.5 there would be 8 rectangles and the sum of their areas would be: 0.5[ f (1) + f (1.5) + f (2) + f (2.5) + f (3) + f (3.5) + f (4) + f (4.5)] square units. From the diagram it can be seen that the left rectangle approximation is less than the actual area under the curve. Worked examPle 18
Find an approximation for the area between the curve f (x) shown and the x-axis from x = 1 to x = 3 using left rectangles of width 0.5 units. The graph shown has the equation f (x) = 0.2x2 + 3.
eBook plus
y
f(x)
Tutorial
int-0565 Worked example 18
3
0 0.5 1 1.5 2 2.5 3 x ThInk
WrITe
Method 1: Technology-free 1
Write the number of rectangles and their width.
There are 4 rectangles of width 0.5 units.
Chapter 9
Integration
435
2
Find the height of each rectangle (left) by substituting the appropriate x-value into the f (x) equation.
h1 = f (1) = 0.2(1)2 + 3 = 3.2 h2 = f (1.5) = 0.2(1.5)2 + 3 = 3.45 h3 = f (2) = 0.2(2)2 + 3 = 3.8 h4 = f (2.5) = 0.2(2.5)2 + 3 = 4.25
3
Area equals the width multiplied by the sum of the heights.
Area = width × (sum of heights of 4 rectangles) = 0.5(3.2 + 3.45 + 3.8 + 4.25) = 0.5(14.7)
4
Calculate this area.
5
State the solution.
The approximate area is 7.35 square units.
= 7.35
Method 2: Technology-enabled 1
On the Main page type: 0.2x2 + 3 Highlight the equation and tap: • Interactive • Define • OK
2
To find the approximate area, complete the entry line as: 0.5( f(1) + f(1.5) + f(2) + f(2.5) Press E.
3
Write the answer with the correct units for area.
∴ The approximate area is 7.35 square units.
The right rectangle method y f(x) Consider the area between the curve f (x) shown at right, the x-axis and the lines x = 1 and x = 5. If the area is approximated by ‘right rectangles’ of width 1 unit then the top right corner of each rectangle touches the curve at one point. So, the height of R1 is f (2) units and the area of R1 is 1 × f (2) square units. R1 R2 R3 R4 Similarly, the area of R2 = 1 × f (3) square units, 0 1 2 3 4 5 x the area of R3 = 1 × f (4) square units the area of R4 = 1 × f (5) square units. Therefore, the approximate area between the curve f (x), the x-axis and the lines x = 1 to x = 5 is (R1 + R2 + R3 + R4) = 1[ f (2) + f (3) + f (4) + f (5)] square units. If the same area was approximated with upper rectangle widths of 0.5 units, the sum of their areas would equal: 0.5[ f (1.5) + f (2) + f (2.5) + f (3) + f (3.5) + f (4) + f (4.5) + f (5)] square units. From the diagram it can be seen that the right rectangle approximation is greater than the actual area under the curve.
For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation.
436
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
Worked Example 19
Find an approximation for the area in the diagram in worked example 18 using right rectangles of width 0.5 units. f (x) = 0.2x2 + 3 Think
Write
1
Find the number of rectangles and the height of each one (from left to right).
There are 4 rectangles: h1 = f (1.5) = 0.2(1.5)2 + 3 = 3.45 h2 = f (2) = 0.2(2)2 + 3 = 3.8 h3 = f (2.5) = 0.2(2.5)2 + 3 = 4.25 h4 = f (3) = 0.2(3)2 + 3 = 4.8
2
Area is the width of the interval multiplied by the sum of the heights.
Area = 0.5(3.45 + 3.8 + 4.25 + 4.8)
3
Calculate the area.
4
State the solution.
The approximate area is 8.15 square units.
= 0.5(16.3) = 8.15
It can be seen that the left rectangle approximation (7.35 units) is less than the right rectangle approximation (8.15 units). If the area is divided into narrower strips, the estimate of the area would be closer to the true value. Worked Example 20
With width intervals of 1 unit, calculate an approximation for the area between the graph of f (x) = x2 + 2 and the x-axis from x = −2 to x = 3 using: a left rectangles b right rectangles c averaging of the left and right rectangle areas. Think
Write
1
Sketch the graph of f (x) over a domain which exceeds the width of the required area
2
Draw the left and right rectangles.
y
y = x2 + 2
2 −2 −1
0 1 2 3 x = Left rectangles = Right rectangles
a
1
2
Calculate the height of the left rectangles by substituting the appropriate values of x into the equation for f (x). Note the two rectangles to the right and left of the origin have the same height and are equal in area. Find the area by multiplying the width by the sum of the heights.
a Left rectangle heights:
f (−2) = (−2)2 + 2 = 6 f (−1) = (−1)2 + 2 = 3 f (0) = 02 + 2 = 2 f (1) = 12 + 2 = 3 f (2) = 22 + 2 = 6
Area = 1(6 + 3 + 2 + 3 + 6) = 20 Using left rectangles, the approximate area is 20 square units.
Chapter 9 Integration
437
b
1
2
Calculate the height of the right rectangles by substituting the appropriate values of x into the equation for f (x).
b Right rectangle heights:
Find the area by multiplying the width by the sum of the heights.
Area = 1(3 + 2 + 3 + 6 + 11) = 25 Using right rectangles, the approximate area is 25 square units. 20 + 25 c Average of the areas = 2 = 22.5 The approximate area is 22.5 square units when averaging the left and right rectangle areas and using widths of 1 unit.
c Find the average by adding the area of the
left rectangles and right rectangles and dividing by 2.
f (−1) = 3 (from above) f (0) = 2 f (1) = 3 f (2) = 6 f (3) = 32 + 2 = 11
Note that this average is between the area of the left rectangles and the area of the right rectangles and is closer to the actual area. rememBer
1. An approximation to the area between a curve and the x-axis can be found by dividing the area into a series of rectangles that are all the same width. The approximation is found by finding the sum of all the areas of the rectangles. 2. For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. 3. For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation. exerCISe
9d
approximating areas enclosed by functions 1 We18 Find an approximation for the area between the curve f (x) at right and the x-axis from x = 1 to x = 5 using left rectangles of width 2 units. 2 Find an approximation for the area between the curves below and the x-axis, from x = 1 to x = 5, by calculating the area of the shaded rectangles. y a b y f(x) (5, 4)
4 2
19
(1, 2)
12 0
438
1
5
x
f(x)
(4, 19) (3, 12)
7 (2, 7) 4 (1, 4) 3 0 1 2 3 4 5x
maths Quest 12 mathematical methods CaS for the Casio ClassPad
y
f(x)
(3, 3) 3 (1, 2) 2 0 1
3
5
x
3 MC Consider the graph of y = x2 from x = 0 to x = 4 at right. a The width of each rectangle is: A 1 unit B 2 units C 3 units D 4 units E varying b The height of the right-most rectangle is: A 9 units B 4 units C 16 units D 12 units E 1 unit c The area between the curve y = x2 and the x-axis from x = 0 to x = 4 can be approximated by the area of the left rectangles as: A 20 sq. units B 14 sq. units C 18 sq. units D 15 sq. units E 30 sq. units 4 WE 19 a Find an approximation for the area in the diagram at right using right rectangles of width 1 unit. b A better approximation for the area under this curve can be found by averaging the right and left rectangle areas. State this approximate value. 5 Find an approximation for the area between the curves below and the x-axis, from x = 1 to x = 5, by calculating the area of the shaded rectangles. b y c a y (1, 8)
8
8 7
6
0
(1, 8) (3, 8)
5 f(x)
1
x
(5, 5)
5
6 4
0 1
g
5
3
1 2 3 4
y 11 10
1 2 3 4
(2, 11)
(3, 10) (4, 7)
x
1 2 3 4 5
f y 4
f(x)
(1, 4)
f(x)
(2, 6)
(3, 2)
2
(1, 4)
x 0
x
f(x)
(4, 10) (3, 9)
x
y = x2
y
x 5 f(x)
y 10 9
f(x)
(3, 3)
3
0
8 7
0 1 3
e
y
y = x2
0
0
d
y
1
2 3 4 5
x
0
3
1
x
5
y 7
(1, 7) (2, 5)
5 4
0
(4, 7)
1
2
3
4
5
f(x) x
6 WE20 With width intervals of 1 unit calculate an approximation for the area between the graph of f (x) = x2 + 4 and the x-axis from x = 1 to x = 4 using: a left rectangles b right rectangles c averaging of the left and right rectangle areas.
y y = x2 + 4
0
1
2 3 4
x
Chapter 9 Integration
439
7 Find the approximate area between the curves below and the x-axis, over the interval indicated, by calculating the area of the shaded rectangles. Give exact answers. y a y b y = ex y = −x2 + 3x + 8
0
1 2 3 4 x = 1 to x = 4
c y
x −1 0 1 2 x = −1 to x = 2
x
y = loge(x)
d y
0
0 1 2 3 4 5 x x = 1 to x = 5
e y
f(x) =
1– 3 3x
0 1 2 3 4 5 x = 1 to x = 5
g
y
− 3x2 + 8x
x
f
y = (x − 4)2
1 2 3 4 5 6 x = 2 to x = 6
f(x) = −x2 − 4x
−3 −2.5 −2 −1.5 −1
x
y
0
x
x = −3 to x = −1
x = 2 to x = 6 2 3 4 5 6 x
0
y = x3 − 6x2
8 Calculate an approximation for the area between the graph of y = x(4 − x), the x-axis and the lines x = 1 and x = 4, using interval widths of 1 unit and: a left rectangles b right rectangles c averaging the left and right rectangle areas. 9 Calculate an approximation for the area under the graph of y = x2 − 4x + 5 to the x-axis between x = 0 and x = 3, using interval widths of 0.5 units and: a left rectangles b right rectangles c averaging the left and right rectangle areas. 10 Find an approximation for the area under the graph of y = 2x between x = 0 and x = 3, using interval widths of 1 unit, by averaging the left and right rectangle areas.
440
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
9e
The fundamental theorem of integral calculus Consider the region under the curve f (x) between x = a and x = b, where f (x) ≥ 0 and is continuous for all x ∈ [a, b]. Let F (x) be the function that is the measure of the area under the curve between a and x. F (x + h) is the area under the curve between a and x + h and F (x + h) − F (x) is the area of the strip indicated on the graph. The area of the strip is between the areas of the left and right rectangles; that is, f (x)h < F (x + h) − F (x) < F (x + h)h or f ( x ) <
F ( x + h) − F F(( x ) < f ( x + h), h ≠ 0 (dividing by h). h
y
0
y = f(x)
a x x+h b F(x) F(x + h) − F(x)
x
As h → 0, f (x + h) → f (x) F ( x + h) − F ( x ) = f ( x) h that is, F ′(x) = f (x) (differentiation from first principles). lim
or
h→0
F ( x ) = ∫ f ( x ) dx
Therefore, that is, F (x) is an antiderivative of f (x)
∫ f ( x) dx = F ( x) + c
or but when x = a,
∫ f ( x) dx = F (a) + c = 0 (as the area defined is zero at x = a) c = −F F((a).
or Therefore,
∫ f ( x) dx = F ( x) − F (a)
and when x = b,
∫ f ( x) dx = F (b) − F (a).
That is, the area under the graph of f (x) between x = a and x = b is F (b) − F (a).
∫ f ( x) dx is the indefinite integral, which represents the general antiderivative of the function
being integrated. This is the fundamental theorem of integral calculus and it enables areas under graphs to be calculated exactly. It applies only to functions that are smooth and continuous over the interval [a, b]. b
It can be stated as area = ∫a f ( x ) dx = [ F ( x )]ba [do not add c as F (x) is an antiderivative of f (x)] = F (b) − F (a) a and b are called the terminals of this definite integral and indicate the domain over which the integral is taken.
Chapter 9
Integration
441
b
∫a
f ( x ) dx is called the definite integral because it can be expressed in terms of its terminals a and b, which are usually real numbers. In this case the value of the definite integral evaluates asis a real number and not a function. The function being integrated, f (x), is called the integrand.
Properties of definite integrals
eBook plus
Definite integrals have the following five properties. 1.
a
∫a
b
Digital doc
f ( x ) dx = 0
Investigation Definite integrals c
b
a
c
2.
∫a
f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx, a < c < b
3.
∫a k f ( x) dx = k ∫a
4.
∫a [ f ( x) + g( x)] dx = ∫a
5.
∫a
b
b
f ( x ) dx
b
b
b
f ( x ) dx =
−
a
∫b
b
f ( x ) dx + ∫ g( x ) dx a
f ( x ) dx
Worked examPle 21
eBook plus
Evaluate the following definite integrals. 3
a
∫0
b
∫1 (2 x + 1)3 dx
Tutorial
int-0566
( 3 x 2 + 4 x − 1) dx
2
Worked example 21
4
ThInk
WrITe
Method 1: Technology-free a
b
1
Antidifferentiate each term of the integrand and write in the form [ F ( x )]ba.
a
3
∫0 (3x 2 + 4 x − 1) dx = [ x 3 + 2 x 2 − x ]30
2
Substitute values of a and b into F (b) − F (a).
= [33 + 2(3)2 − 3] − [0 3 + 2(0)2 − 0] =
3
Evaluate the integral.
= 42 − 0 = 42
1
Express the integrand with a negative power.
b
2
4
2
−
∫1 (2 x + 1)3 dx = ∫1 4(2 x + 1) 3 dx 2
2
Antidifferentiate by rule.
4(2 x + 1) − 2 = 2 × − 2 1
2
− = − (2 x + 1) 2 1
3
442
Express the integral with a positive power.
maths Quest 12 mathematical methods CaS for the Casio ClassPad
2
− 1 = 2 (2 x + 1) 1
2
−1 −1 = 2 − 2 5 3 −1 1 = + 25 9 16 = 225
Evaluate the definite integral.
5
−1 = 2 (2 x + 1) 1
Substitute the values of a and b into F (b) − F (a) where a = 1 and b = 2.
4
Method 2: Technology-enabled 1
On the Main page, using the soft keyboard, tap: • ) • • P Complete the entry lines as: 3
∫0 (3x 2 + 4 x − 1) dx 2
4
∫1 (2 x + 1)3 dx Press E after each entry.
2
Write the answers.
3
a
∫0 (3x 2 + 4 x − 1) dx = 42
b
∫1 (2 x + 1)3 dx = 225
2
4
16
Worked Example 22
Find the exact value of each of the following definite integrals. a
2π
∫π
x sin dx 6
b
∫0 ( e 3 x − e 1
−3x
) dx
Think a
Write
1
Antidifferentiate the integrand, writing it in the form [ F ( x )]ba.
2
Substitute values of a and b into F (b) − F (a).
a
2π
∫π
2π
x x sin dx = − 6 cos 6 6 π
− 2π − π = 6 cos 6 − 6 cos 6 − π − π = 6 cos − 6 cos 3 6
Chapter 9 Integration
443
3
− 1 − 3 = 6 2 − 6 2
Evaluate the integral.
= [− 3] − [− 3 3 ] = −3+ 3 3 b
1
Antidifferentiate the integrand, using [ F ( x )]ba.
2
Substitute values of a and b into F (a) − F (b).
3
Evaluate.
b
1
∫0 (e3 x − e
−3 x
) dx 1
− = 13 e3 x + 13 e 3 x 0 − = 13 e3 + 13 e 3 − 13 e 0 + 13 e 0
= 13 e3 + 13 e = 13 ( e3 + e
−3
−3
− 23
− 2)
Worked Example 23
If
k
∫0 8 x dx = 36, find k.
Think
Write
Method 1: Technology-free 1
2
Antidifferentiate the integrand, using [ F ( x )]ba.
3
Substitute the values of a and b into F (a) − F (b). Simplify the integral.
4
Solve the equation.
k
∫0 8x dx = [4 x 2 ]0k So [4 x 2 ]k0 = 36 [4k2] − [4(0)2] = 36 4k2 − 0 = 36 4k2 = 36 k2 = 9 k=± 9 k = 3 or −3
Method 2: Technology-enabled 1 On the Main page, complete the entry line as: k
∫ 0 8x dx = 36
Highlight the integral and tap: • Interactive • Equation/Inequality • solve Change the variable to k. • OK
2
444
Write the solutions.
k
Solving ∫ (8 x ) dx = 36 for k implies k = −3 or k = 3. 0
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
Sigma notation
y
An alternative notation for the definite integral of f (x) for x ∈[a, b] is the sigma notation (meaning ‘the sum of’). y = f(x)
(xi, f(xi))
a
x
b
x
Divide the interval [a, b] into n equal subintervals with the ith subinterval of width δxi and height f (xi), i ∈[1, n]. The area of the rectangle formed by the ith subinterval is δAi = f (xi) × δxi. n
b
As n → ∞, δxi → 0 and ∑ δ Ai → ∫ f ( x ) dx . a
1
So
n
b
∑ f ( xi )δ xi δ x→0
∫a
f ( x ) dx = lim
i =1
Worked Example 24
The interval [1, 3] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 1 = x0 < x1 < . . . < xn − 1 < xn = 3. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n
∑ ( 6 x i 2δ x ) . δ x→ 0
Rewrite as a definite integral and thus evaluate lim Think 1
2
i=1
Write n
Recognise the alternate notation for the definite integral.
∑ (6 xi 2δ x) = ∫ 1 6 x 2 dx δ x→0
Evaluate.
∫ 1 6 x 2 dx = 2 x3 1
3
lim
i =1
3
3
= 2 × 27 − 2 × 1 = 54 − 2 = 52
REMEMBER
1. The fundamental theorem of calculus is where F (x) is an antiderivative of f (x). 2. The expression
b
∫a
b
∫a
f ( x ) dx = [ F ( x )]ba = F (b) − F (a).
f ( x ) dx is called the definite integral where a and b are the
terminals and represent the upper and lower values of x. 3. Properties of definite integrals a
(a) ∫ f ( x ) dx = 0 a
b
c
b
a
a
c
(b) ∫ f ( x ) dx = ∫ f ( x )dx + ∫ f ( x ) dx, a < c < b
Chapter 9 Integration
445
4.
exerCISe
9e
b
b
(c)
∫a k f ( x) dx = k ∫a
(d)
∫a [ f ( x) + g( x)] dx = ∫a
(e)
∫a
b
∫a
b
b
b
Digital doc
SkillSHEET 9.2 Subtracting function values
a
∫b
b
f ( x )dx + ∫ g( x ) dx a
f ( x ) dx
n
f ( x ) dx = lim
δ xi → 0
∑ f ( xi ) δ xi i =1
The fundamental theorem of integral calculus Evaluate the following definite integrals.
1
a
∫0 x 2 dx
d
∫2 x 2 dx
b
1
6
1
e
g
∫ −1
j
∫1 (4 x −2 + 2 x − 6) dx
(6 − 2 x + x 2 ) dx
2
0
m
∫−2
p
∫1
s
∫−2
− 4 ( 2 − 3 x )3
3 2x3
6
0
2
dx
+ 5x 2 dx x
8 − 3x
2 We22
446
−
f ( x ) dx =
1 We21 eBook plus
f ( x ) dx
∫0
d
∫1 (3e6 x + 5x) dx
g
∫32 sin ( x) dx
j
∫π
2
π
− 2 sin
( 3x ) dx
x 3 cos dx 6
m
∫π
p
∫0 π 2 + sin
π
∫0
( x 3 + 3 x 2 − 2 x ) dx
−2
h
∫− 4 ( x3 + x − 4) dx
k
∫0 2( x + 4)4 dx
3
3
5 dx (2 x − 7)3
n
∫0
q
∫1 5x dx
3
5
∫3 ( x 2 − 2 x) dx
f
∫1 (3x 2 + 2 x 2 ) dx
i
∫4 3
l
∫ 1 3(5x − 2)4 dx
o
∫1 2 x 2 − 3x
r
∫3
1
4
9
x dx
2
−
4
3
7
1 2x − 5
−1
dx
dx
Find the exact value of each of the following definite integrals.
e 4 x dx
2π
2
4
c
dx
a
2π
3
∫0 x3 dx
( 4x ) dx
0
x
b
∫− 2
e
∫1 x + e 2 dx
h
∫π2 3 sin (4 x) dx
k
∫− π cos (2 x) dx
4
e 3 dx 5
x
π
0
0
x
n
∫− 2π − 7 cos 2 dx
q
∫1 [ x 2 + 3 − 6 sin (3x)] dx
3
maths Quest 12 mathematical methods CaS for the Casio ClassPad
1
−
c
∫−1 − 4e 2 x dx
f
∫−3 (e2 x − e 2 x ) dx
i
∫0 5 sin
l
∫−2π 8 cos (4 x) dx
−1
π
−
( 4x ) dx
π
2
−π 2 −4 −π
o
∫
r
∫1 x + 3 cos
π
1
cos (3 x ) dx
( 2x ) dx
3 If
5
∫1
f ( x ) dx = 6, find the value of
4 MC Given that a
5
∫ 1 f ( x) dx = 6,
5
∫ 1 [ f ( x) + 1] dx is equal to: A 16
b
5
∫1 3 f ( x) dx.
B 10
C 11
D 19
E 22
C −5
D 6
E 0
1
∫5 f ( x) dx is equal to: A −6
B 5
5 Evaluate the following. 3
a
∫0
d
∫5π
g
∫1 t
(t 2 − 4t ) dt
10π
− sin x dx 5
4 −3
t
dt
π 2 0
2 cos (3t ) dt
π
b
∫
e
∫0 e 4 − cos (2 x) dx
h
∫−1 [3 sin (2 x) − e
x
1
−3x
3
7
c
∫4
f
∫1 4m − 3 dm
2
3 ( p − 3) 2
dp
8
] dx
k
6 WE23 If ∫ (2 x + 3) dx = 4, find k. 0
k
7 If ∫ 3 x 2 dx = 8, find k. 0
8 If ∫
k 1
2 dx = log e (9), find k. x x
a
9 If ∫ e 2 dx = 4, find the value of a. 0
π
10 If ∫ cos (2 x) dx = − k
π 3 , find the value of k given that 0 < k < . 2 4
6
11 Given that ∫ f ( x ) dx = 8, evaluate the following. 2
a
6
∫ 2 2 f ( x) dx
b
2
∫6
f ( x ) dx c
6
∫2
[3 f ( x ) − 3] dx d
6
∫2 [ f ( x) + 2 x] dx
12 WE24 MC The interval [1, 3] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 1 = x0 < x1 < . . . < xn − 1 < xn = 3. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n
Then lim
δ x→0
A
1
∫3
∑ ( xi3 δ x) is equal to: i =1
x 3 dx b ∫
3 1
3 x4 dx c x 3 D 20 E 26 1 4
13 MC The interval [0, 3] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 0 = x0 < x1 < . . . < xn − 1 < xn = 3. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n
∑ ( xi 2δ x) is equal to: δ x→0
Then lim
i =1
A 9 b
3
∫0
3 x3 dx c x 3 0 D 3
0
∫ 3 x 2 dx
E 27
Chapter 9 Integration
447
9f
Signed areas When calculating areas between the graph of a function f (x) and the x-axis using the definite integral
b
∫a
f ( x ) dx, the area is signed; that is, it is positive or negative. If f (x) > 0, the region is
above the x-axis; if f (x) < 0 it is below the axis. We shall now examine these two situations and look at how we calculate the area of regions that include both.
Region above axis
y
If f (x) > 0, that is, the region is above the x-axis, then b
∫a
y = f(x)
f ( x ) dx > 0, so the value of the definite integral is positive. For example, if f (x) > 0, then the area =
b
∫a
f ( x ) dx. 0
Region below axis
a
b
x
If f (x) < 0, that is, the region is below the x-axis, then b
∫a
f ( x ) dx < 0, so the value of the definite integral is negative. b
For example, if f (x) < 0, then the area = − ∫ f ( x ) dx
y
a
or
b
∫a
f ( x ) dx , as the region is below the x-axis or
y = f(x) a
b x
0
a
area = ∫ f ( x ) dx (reversing the terminals changes the sign). b
Therefore, for areas below the x-axis, ensure that the area has a positive value. (Areas cannot be negative.)
Combining regions For regions that are combinations of areas above and below the x-axis, each area has to be calculated by separate integrals — one for each area above and one for each area below the x-axis. For example, from the diagram, Area = A1 + A2 b
However, ∫ f ( x ) dx = A1 − A2 , because the areas are signed.
y
y = f(x)
A1 a
0 A 2
c
b
a
To overcome this difficulty we find the correct area as: b
c
c
a
Area = ∫ f ( x ) dx − ∫ f ( x ) dx (= A1 − − A2 = A1 + A2 ) b
or = ∫ f ( x ) dx + c
c
∫a f ( x) dx
b
a
c
c
or = ∫ f ( x ) dx + ∫ f ( x ) dx Note: When calculating the area between a curve and the x-axis it is essential that the x-intercepts are determined and a graph of the curve is sketched over the interval required. The term |x| means that we should make the value of x positive even if it is negative. 448
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
x
Worked Example 25 a Express the shaded area as a definite integral. b Evaluate the definite integral to find the shaded area, giving your
y
y = –x1
answer as an exact value.
0 Think
1
4
x
Write 4
∫1
1 dx x
a
Express the area in definite integral notation where a = 1 and b = 4.
a Area =
b
1
Antidifferentiate the integrand.
b Area = log e x 1
2
Evaluate.
3
State the solution as an exact answer.
4
= [loge (4)] – [loge (1)] = loge (4) − 0 = loge (4) The area is loge (4) square units.
Worked Example 26
Calculate the shaded area.
y
y = x2 − 4x
0
Think 1
Express the area in definite integral notation, with a negative sign in front of the integral as the region is below the x-axis.
2
Antidifferentiate the integrand.
3
Evaluate.
4
State the solution.
1
3 4
x
Write 3
Area = − ∫ ( x 2 − 4 x ) dx 1
= − [ 1 x 3 − 2 x 2 ]13 3
= − [( 3 (3)3 − 2(3)2 ) − ( 3 (1)3 − 2(1)2 )]
1 − = [(9 − 18) − ( 3 − 2)]
− − − 2 = [ 9 − ( 1 3 )]
= − [− 9 + 1 23 ]
= − (− 7 13 )
= 7 13
1
1
1
The area is 7 3 square units.
Chapter 9 Integration
449
Worked examPle 27 y
a Express the shaded area as a definite integral. b Calculate the area. c Calculate the area using a CAS calculator.
0
−2 ThInk
y = x3 − 4x 2
WrITe 0
2
∫−2 ( x3 − 4 x ) dx − ∫0 ( x3 − 4 x ) dx
a
Express the area above the x-axis as an integral and the area below the x-axis as an integral. For the area below the x-axis, take the negative of the integral from 0 to 2.
a Area =
b
1
Antidifferentiate the integrands.
b = 14 x 4 − 2 x 2 − − 14 x 4 − 2 x 2 2 0
2
Evaluate.
=
3
Simplify.
4
State the solution. On the Main page, using the soft keyboard, tap: • ) • • P Complete the entry line as:
= [0 − (4 − 8)] − [(4 − 8) − 0] = 4 − (−4) =8 The area is 8 square units.
c
1
x
0
0
2
( 14 (0)4 − 2(0)2 ) − ( 14 (−2)4 − 2(−2)2 ) − ( 14 (2)4 − 2(2)2 ) − ( 14 (0)4 − 2(0)2 )
c
2
∫ − 2 ( x3 − 4 x) dx − ∫ 0 ( x3 − 4 x) dx Press E. Alternatively, take the absolute value for the area below the x-axis as shown. 2
Write the area with the correct units.
∴ The area is 8 square units.
Worked examPle 28
eBook plus
a Sketch the graph of y = ex − 2 showing all intercepts and using exact values
for all key features. b Find the area between the curve and the x-axis from x = 0 to x = 2. ThInk a
450
1
WrITe
Find the x-intercept by letting y = 0 and solving for x.
a When y = 0, ex – 2 = 0
maths Quest 12 mathematical methods CaS for the Casio ClassPad
ex = 2
Tutorial
int-0567 Worked example 28
loge (ex) = loge (2) x = loge (2) (or approximately 0.69) so the x-intercept is loge (2). When x = 0, y = e0 − 2 =1−2 = −1 so the y-intercept is −1.
Take loge of both sides.
b
2
Find the y-intercept by letting x = 0.
3
Note the vertical translation and hence sketch the graph showing the appropriate horizontal asymptote and intercept. Shade the region required.
1
Express the area above the x-axis as an integral and the area below the x-axis as an integral. Subtract the area below the x-axis from the area above the x-axis. Antidifferentiate the integrands.
3
Evaluate. (Remember: eloge (a) = a)
4
Simplify.
5
State the solution.
0 −1
−2
4
2
y y = ex − 2
−2
b Area =
loge2
x 2 y=−2
2
log e ( 2)
∫loge (2) (e x − 2) dx - ∫0
= [e x − 2 x ]2log
(e x − 2) dx
log e ( 2)
e ( 2)
− [e x − 2 x ]0
= [e2 − 2(2)] − [eloge (2) − 2 loge (2)] − [[eloge (2) − 2 (loge (2)] − [e0 − 2(0)]] = [e2 − 4] − [2 − 2 loge (2)] − [[2 − 2 loge (2)] − [1 − 0]] = e2 − 4 − 2 + 2 loge (2) − 2 + 2 loge (2) + 1 = e2 − 7 + 4 loge (2) The area is [e2 − 7 + 4 loge (2)] or approximately 3.162 square units.
REMEMBER
1. If f (x) > 0, area =
b
∫a
f ( x ) dx .
b
2. If f (x) < 0, area = −∫ f ( x ) dx , as the region is below the x-axis or a
b
∫a
=
= ∫ f ( x ) dx , (reversing the terminals changes the sign).
f ( x ) dx , or
a
b
y
3. Check if the required area lies above and below the x-axis. 4. Area =
b
∫c
f ( x ) dx − ∫ f ( x ) dx (= A1 − −A2 = A1 + A2)
b
or = ∫ f ( x ) dx + c
5. e
log e ( a )
y = f(x)
c
a
A1
c
∫a f ( x) dx
a
0 A 2
c
b
x
=a
Chapter 9 Integration
451
Exercise
9f
Signed areas y
1 Find the area of the triangle at right. a geometrically b using integration
y=x
0
x
4
y
2 Find the area of the triangle at right. a geometrically b using integration
3 0
3 y=3−x
x
3 WE25a Express the following shaded areas as definite integrals. a y b y c y = 2x
y
y = x2
4
01 y
d
−3
y = 3x2
y
e
x
−1 0 y
g
0
x
3
0
y = ex
j
x y=4–x
y = x3 − 9x2 + 20x
1
0
3
x
1 2 y
f
0
−2
x
h y
x
y = −x3 − 4x2 − 4x
i
y 2
y = 2 sin (2x)
−2x
y=e
1 0
−1 y
4
x
1
0
0
1
4
π – 2
x
x
y = cos (–3x)
1 0
x
3— π 2
4 WE25b Evaluate each of the definite integrals in question 3 to find the shaded area. Give your answer as an exact value. 5 For each of the following, sketch a graph to illustrate the region for which the definite integral gives the area.
452
3
a
∫0 4 x dx
d
∫−1 (4 − x 2 ) dx
g
∫2 loge
1
4
( 2x ) dx
2
b
∫1 (6 − x) dx
e
∫1
h
4
x dx
π 2 3 sin (2 x ) dx 0
∫
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
3
c
∫1 x 2 dx
f
∫−3 2e x dx
0
6 WE26 Calculate each of the shaded areas below. y y a b − y=x−2
0
y y = x2 − 4
c
x
−1 0
−2
x
2
y = 4 − 2x
0
2
−2 y
d −2
−1
x
0
y
e
y = x3
y −1
y = x3 + 2x2 − x −2
0
h
x
1
0 1 x
−1
y = 1− x2
g
y
f
x
0
−2
x
y
1– 2
0
1
i
y y = sin (x)
x 0
y=
y = −ex
π
−e−2x
3π — 2
2π
x
y
j
−2π
x
−π 0
y = 2 cos (–2x )
7 MC a The area between the graph, the x-axis and the lines x = −2 and x = −1 is equal to: 2
A
∫1
C
∫− 1 f ( x) dx
f ( x ) dx
0
y
−1
B
∫− 2 f ( x) dx
D
∫2 f ( x) dx
1
0
−2
−1
3 y = f(x)
x
E − ∫ − f ( x ) dx 2 b The area between the graph, the x-axis and the lines x = −2 and x = 3 is equal to: 3
−2
1
3
0
3
A
∫0 f ( x) dx + ∫0
C
∫− 2 f ( x) dx + ∫1
E
∫− 2 f ( x) dx − ∫0 f ( x) dx
f ( x) dx
B
f ( x) dx
D
3
∫− 2 f ( x) dx −2
∫3
f ( x ) dx
8 WE27a Express the following shaded areas as definite integrals which give the correct area. b c a y y y g(x) f(x) 0
2
5
x
−3
0 1
3
x
−3
−1
0 2
x h(x)
Chapter 9 Integration
453
y
d
−5 −4 −2
y
e
0
f(x)
x
g(x)
0
−3 −2
2 3
x
9 mC Examine the graph. y
0
−2
1
3
x
y = x3 − 2x2 − 5x + 6
a The area between the curve and the x-axis from x = −2 and x = 1 is equal to: 1
B 154 sq. units
3
E 10 sq. units
A 1712 sq. units D −154 sq. units
3
1
C −1712 sq. units
b The area between the curve and the x-axis from x = 1 and x = 3 is equal to: A −6 2 sq. units B 2 sq. units C −513 sq. units 3
D 513 sq. units
E 6 2 sq. units 3
c The area between the curve and the x-axis from x = −2 and x = 3 is equal to: 5
3
A 1012 sq. units
B 114 sq. units
D 12 sq. units
E 2112 sq. units
5
C 2212 sq. units
1
10 We28 Sketch the graph of the curve y = x2 − 4, showing all intercepts and using exact values for all key features. Find the area between the curve and the x-axis: a from x = 0 to x = 2 b from x = 2 to x = 4 c from x = 0 to x = 4. 11 Sketch the graph of the curve y = x3 + x2 − 2x, showing all intercepts. Find the area between the curve and the x-axis between the lines: a x = −2 and x = 0 b x = 0 and x = 1 c x = −2 and x = 1. 12 Sketch the graph of the curve y = 1 + 3 cos (2x) over [0, π]. Find the exact area between the curve and the x-axis from: π a x = 0 to x = 4 3π b x= to x = π. 4 eBook plus Digital doc
WorkSHEET 9.2
454
13 Sketch the graph of f (x) = x − 1 and find the area between the curve and the x-axis and the lines x = 2 and x = 3. Give both an exact answer and an approximation to 3 decimal places. 14 Find the exact area between the curve y = 1 , the x-axis and the lines x = 12 and x = 2. x 15 Find the exact area bounded by the curve g(x) = ex + 2, the x-axis and the lines x = −1 and x = 3.
maths Quest 12 mathematical methods CaS for the Casio ClassPad
9g
Further areas Areas bound by a curve and the x-axis For graphs with two or more x-intercepts, there is an enclosed region (or regions) between the graph and the x-axis. The area bound by the graph of f (x) and the x-axis is: −
b
∫a
or
y
f ( x ) dx (negative because the area is below the x-axis) b
∫a
y = f(x)
0
a
x
b
f ( x ) dx . y
The area bound by the graph of g(x) and the x-axis is: b
c
∫c g( x) dx − ∫a g( x) dx or
c
b
∫a g( x) dx + ∫c g( x) dx
a
c
0
b
x
That is, if the graph has two x-intercepts then one integrand is y = g(x) required. If the graph has three x-intercepts then two integrands are required, and so on. Note: Wherever possible it is good practice to use sketch graphs to assist in any problems involving the calculation of areas under curves. Worked Example 29 a Sketch the graph of the function g(x) = (3 − x)(2 + x). b Find the area bound by the x-axis and the graph of the function. Think a
b
Write
1
Determine the type of graph by looking at the number of brackets and the sign of the x-terms.
2
Solve g(x) = 0 to find the x-intercepts.
3
Sketch the graph.
1
Shade the region bound by g(x) and the x-axis.
a g(x) = (3 − x)(2 + x) is an inverted parabola.
For x-intercepts, g(x) = 0 (3 − x)(2 + x) = 0 x = 3 and x = −2. The x-intercepts are −2 and 3. b
y
x 3 y = g(x)
−2 0
3
∫− 2 (6 + x − x 2 ) dx
2
Express the area as an integral.
Area =
3
Evaluate.
= [6 x + 12 x 2 − 13 x 3 ]3− 2
= [6(3) + 2 (3)2 − 3 (3)3]
1
1
1
1
−[6(−2) + 2 (−2)2 − 3 (−2)3]
Chapter 9 Integration
455
4
9
8
= (18 + 2 − 9) − (−12 + 2 + 3 )
= 13 2 − (−7 3 )
= 13 2 + 7 3
= 20 6
1 1
1
1
5 5
The area bound by g(x) and the x-axis is 20 6 square units.
State the solution.
Finding areas without sketch graphs When finding areas under curves that involve functions whose graphs are not easily sketched, the area can be calculated providing the x-intercepts can be determined. Note: The symbol f (x)is known as the absolute value of f (x), which means that we must make the f (x) function or value positive whenever it is negative. For example, −5= 5
3= 3
−2 3
2
=3
As areas cannot be negative, taking the absolute value of the integrands involved in a problem will ensure that all areas are made positive. y For example, The shaded area at right =
or
=
b
∫c
b
∫c
f ( x ) dx + f ( x ) dx +
c
∫a f ( x) dx
y = f(x) a
c 0
b
c
∫a f ( x) dx
Worked Example 30 a Find the x-intercepts of y = sin (2x) over the domain [0, 2π]. b Calculate the area between the curve, the x-axis and x = 0 and x = π. Think a
Write
1
To find the x-intercepts, let y = 0.
2
Solve for x over the given domain.
a For x-intercepts, y = 0
sin (2x) = 0 2x = 0, π, 2π, 3π, 4π, etc.
b
456
1
Pick the x-intercepts that are between the given end points of the area.
2
State the regions for which it is necessary to calculate the area.
b x=
x = 0,
3π π , π , , 2π 2 2
π is the only x-intercept between 0 and π. 2 π
Area =
π
∫02 sin (2x) dx + ∫π sin (2x) dx
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
2
x
π
3
Evaluate the absolute value of the integral for each region.
π
= [− 12 cos (2x )] 2 + [− 12 cos (2x )]π
0
=
[− 12
cos
2
(π ) − (− 12
cos (0))]
+ [− 12 cos (2π ) − (− 12 cos (π ))]
= [ 12 − (− 12 )] + [− 12 − ( 12 )]
= 1 + −1 =1+1 =2
4
Add the result to give the total area.
5
State the solution.
The area is 2 square units.
Worked Example 31 a Differentiate loge (x2 − 1).
x . −1 x c Find the area between the graph of 2 , the x-axis, x = 2 and x = 3, giving your answer correct x −1 to 2 decimal places. b Hence, find an antiderivative of
x2
Think a
Write a Let y = loge (x2 − 1)
1
Let y equal the expression to be differentiated.
2
Express u as a function of x in order to apply the chain rule for differentiation. (Let u equal the function inside the brackets.)
3
Find
4
Write y in terms of u.
5
Find
dy . du
dy 1 = du u
6
Find
dy using the chain rule. dx
So
dy 1 = × 2x dx u
Let u = x2 − 1
du . dx
du = 2x dx
y = loge (u)
=
b
dy
∫ dx dx = y + c, express the
1
Since
2
relationship in integral notation. dy Remove a factor from so that it dx resembles the integral required.
b
2x
2x −1
x2
∫ x 2 − 1 dx = loge 2∫
x2
x2 − 1 + c
x dx = log e x 2 − 1 + c −1
Chapter 9 Integration
457
3
∫
Divide both sides by the factor in order to obtain the required integral.
x dx = 12 log e x 2 − 1 + c x2 − 1
An antiderivative of c
1
Find the x-intercepts. x (For 2 = 0, the numerator = 0.) x −1
2
If the x-intercepts are not between the terminals of the area, find the area by evaluating the integrand.
c For x-intercepts,
Area =
State the solution.
x 1 is 2 log x2 − 1. −1
x =0 x2 − 1 x=0
x
∫2 x 2 − 1 dx 3
1 2 = 2 log e x − 1 2
= 1 log e (32 − 1) − 1 log e (22 − 1) 2 2
=
=
3
3
x2
1 2
log e (8) − 12 log e (3)
() = 12 log e ( 83 ) 1 log e 83 2
The area is 12 log e square units.
( 83 ) or approximately 0.49
REMEMBER
1. For graphs with two or more intercepts, there is an enclosed region (or regions) between the graph and the x-axis. 2. The number of regions is one less than the number of intercepts. 3. Where possible, sketch graphs to make it easier to calculate the areas under curves. 4. As areas cannot be negative, take the absolute values of the integrals. 5. When graphs are not easily drawn, areas can be calculated by finding the x-intercepts and determining whether they are within the bounds of the required area. Exercise
9G
Further areas In the following exercise give all answers correct to 2 decimal places where appropriate, unless otherwise stated. 1 WE29 i Sketch the graph of each of the following functions. ii Find the area bound by the x-axis and the graph of each function. a f (x) = x2 − 3x b g(x) = (2 − x)(4 + x) 2 Find the area bound by the x-axis and the graph of each of the following functions. a h(x) = (x + 3)(5 − x) b h(x) = x2 + 5x − 6 2 c g(x) = 8 − x d g(x) = x3 − 4x2 e f (x) = x(x − 2)(x − 3) f f (x) = x3 − 4x2 − 4x + 16 3 2 g g(x) = x + 3x − x − 3 h h(x) = (x − 1)(x + 2)(x + 5) 3 MC The area bound by the curve with equation y = x2 − 6x + 8 and the x-axis is equal to: A 1 13 square units B 6 23 square units C 12 square units D 3 square units
458
E −1 13 square units
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
y
4 MC The area between the curve at right, the x-axis and the lines x = −3 and x = 4 is equal to: A
4
∫− 3 f ( x) dx −3
C
∫4
E
∫− 3 f ( x) dx − ∫2
f ( x ) dx
2
2
4
∫2
D 4
4
∫− 3 f ( x) dx + ∫2
B
f ( x) dx
−3
0 2
x
4 y = f(x)
2
f ( x) dx − ∫ − f ( x ) dx 3
f ( x) dx
5 MC The area between the curve y = x2 − x − 6, the x-axis and the lines x = 2 and x = 4 is equal to: 2 3
A 2 65 square units
B
C 5 square units
D 2 43 square units
E 4 12 square units
square units
6 For each of the following: i sketch the graph of the curve over an appropriate domain, clearly labelling any x-intercepts in the interval required. ii find the area between the curve, the x-axis and the lines indicated below. 2 a y = 3 − 3x2, x = 0 and x = 2 b y = , x = 1 and x = 3 x −1 c y = , x = 1 and x = 2 d y = x3 − 4x, x = −2 and x = 1 x2
e y = e2x, x = −2 and x = 0
g y = 2 sin (x), x =
i y = sin (3x), x =
π π and x = 6 3 −π
− and x = π 2 6
f y = e−x, x = 0 and x = 2
π x h y = cos , x = and x = 2π 2 2 j y = x x, x ≥ 0, x = 0 and x = 4
7 WE30 For each of the following functions: i find the x-intercepts over the given domain ii calculate the area between the curve, the x-axis and the given lines. Use sketch graphs to assist your workings. − a y = x − 4x 1, x ≠ 0, x = 1 and x = 3 b y = sin (x) − cos (x), x ∈ [0, π], x = 0 and x = π c y = e x − e, x = 0 and x = 3 1 d y = x − 2 , x ≠ 0, x = 12 and x = 2 x x
e y = e 2 , x = −2 and x = 2 f y = x4 − 3x2 − 4, x = 1 and x = 4 g y = (x − 2)4, x = 1 and x = 3 8 Find the exact area of the region enclosed by the x-axis, y = e3x and the lines x = 1 and x = 2. π 9 Find the exact area of the region enclosed by the x-axis, y = −cos (x) and the lines x = and 3 5π x= . 6 10 Find the area bound by y = (x − 1)3, the x-axis and the y-axis. 1 11 a Sketch the graph of y = showing all asymptotes and intercepts. ( x − 3)2 b Find the area under the curve between x = −1 and x = 1.
Chapter 9 Integration
459
−
12 a Give the equation of the asymptotes for the function f (x) = (x + 2) 3. b Find the area between the curve, the x-axis, x = −1 and x = 1. 13 Find the area bound by the curve y = 3 − e2x, the x-axis, x = −2 and x = 0. (Find the x-intercepts first.) −π 14 Find the area bound by the curve y = 4 − sin (2x), the x-axis, x = and x = π . (Check the 2 x-intercepts first.) 15 WE31 a Differentiate x loge (x). (x > 0) b Hence, find an antiderivative of loge (x). c Find the area bound by the graph of loge (x), the x-axis, x = 1 and x = 4 giving exact answers. x 16 a Differentiate loge (x2 + 2). b Hence, find an antiderivative of 2 . x +2 x c Find the area between , the x-axis, x = −1 and x = 1. 2 +2 x 17 a Find the area between the graph of y = x2, the x-axis, x = 0 and x = 2. y b Use this result to calculate the area between the graph, the y-axis y = e2x and the line y = 4.
0 2
x
18 Find the exact area of the shaded region on the graph y = e2x below. y
y = x2 (2, 4)
0
2
x
19 Find the shaded area below. (Hint: It is easier if you use symmetry.) y
− π–2
0
π – 2
x
y = 2 sin (x)
20
9H
a The area of the region bounded by the y-axis, the x-axis, the curve y = 2e-x and the 3 line x = k, where k is a positive real constant, is square units. Find k. 2 b The area of the region bounded by the y-axis, the x-axis, the curve y = sin (2x) and the line x = k, where k is a positive real constant, is 1 square unit. Find k.
Areas between two curves We shall now consider the area between two functions, f (x) and g(x), over an interval [a, b]. Our approach depends on whether the curves intersect or do not intersect over this interval.
If the two curves f (x) and g(x) do not intersect over the interval [a, b] Here, we may look at three circumstances: when the region is above the x-axis, when it is below the x-axis, and when it crosses the x-axis.
460
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
y
Region above x-axis b
∫a
The brown shaded area =
b
f ( x ) dx − ∫ g( x ) dx
g(x)
a
b
∫a [ f ( x) − g( x)] dx.
=
f(x)
0 a
x
b
Note: The lower function is subtracted from the higher function to ensure a positive answer. y
Region below x-axis Again, the lower function is subtracted from the higher function to ensure a positive answer. Brown shaded area =
0
a
b
b f(x)
∫a [ f ( x) − g( x)] dx, as f (x) is above g(x)
over the interval [a, b].
y
b
∫a [ f ( x) − g( x)] dx
0
a
g(x)
f(x) g(x)
Region crosses x-axis Shaded area =
x
x
b
Worked Example 32 y
a State the definite integral that describes the shaded area on the
y = 2x + 1
graph at right.
y=x
b Find the area.
0
Think a
x
Write a f (x) = 2x + 1 and g(x) = x
1
State the two functions f (x) and g(x).
2
Subtract the equation of the lower function from the equation of the upper function and simplify.
f (x) − g(x) = 2x + 1 − x =x+1
3
Write as a definite integral between the given values of x.
Area =
b
2
=
2
∫0 [ f ( x) − g( x)] 2
∫0 ( x + 1) dx
b Area = [ 12 x 2 + x ]20
1
Antidifferentiate.
2
Evaluate the integral.
3
State the area.
The area is 4 square units.
= [ 12 (2)2 + 2] − [ 12 (0)2 + 0] = (2 + 2) − (0) =4
Worked Example 33
a Find the values of x where the functions y = x and y = x2 − 2 intersect. b Sketch the graphs on the same axes.
c Hence, find the area bound by the curves. Think a
1
Write/draw
State the two functions.
a y = x and y = x2 − 2
Chapter 9 Integration
461
2
Find where the curves intersect.
3
Solve for x.
For points of intersection: x = x2 − 2 2 x −x−2=0 (x − 2)(x + 1) = 0 x = 2 or x = −1 b For y = x,
b Find the key points of each
y when x = 0, y = 0 y = x2 − 2 when x = 2, y = 2 (2, 2) when x = −1, y = −1 y=x Line passes through (0, 0), (2, 2) and (−1, −1) x 0 (−1, −1) For y = x2 − 2, when x = 0, y = −2 Hence the y-intercept is −2. Parabola also passes through (2, 2) and (−1, −1).
function and sketch.
c
c Let f (x) = x and g(x) = x2 − 2
1
Define f (x) and g (x).
2
Write the area as a definite integral between the values of x at the points of intersection.
Area = ∫ − [ f ( x ) − g( x )] dx
= ∫ − ( x − x 2 + 2) dx
3
Antidifferentiate.
= [ 12 x 2 − 13 x 3 + 2 x ]2−1
4
Evaluate the integral.
= [ 12 (2)2 − 13 (2)3 + 2(2)] − [ 12 (− 1)2 − 13 (− 1)3 + 2(− 1)]
= (2 − 83 + 4) − ( 12 +
= (3 13 ) − (-1 16 )
= 3 13 + 1 16
= 4 12
5
2
1
State the area.
2
= ∫ − [ x − ( x 2 − 2)] dx 1
2
1
1 3
− 2)
The area is 4 12 square units.
If the two curves intersect over the interval [a, b]
y
Where c1 and c2 are the values of x where f (x) and g(x) intersect over the interval [a, b], the area is found by considering the intervals [a, c1], [c1, c2] and [c2, b] separately. For each interval care must be taken to make sure the integrand is the higher function. Subtract the lower function. So the shaded area equals: c1
c2
∫ a [ g( x) − f ( x)] dx + ∫c
1
a c1 0
g(x)
c2
b
x f(x)
b
[ f ( x ) − g( x )] dx + ∫ [ g( x ) − f ( x )] dx c2
Therefore, when finding areas between two curves over an interval, it must be determined whether the curves intersect within that interval. If they do, the area is broken into sub-intervals as shown above. As with areas under curves, sketch graphs should be used to assist in finding areas between curves.
462
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
If sketch graphs are not used, the absolute value of each integral, for each sub-interval, should be taken to ensure the correct value is obtained.
Worked examPle 34
eBook plus
a Find the values of x where the graph of the functions f (x) =
4 and x
Tutorial
int-0801 g(x) = x intersect. Worked example 34 b Sketch the graphs on the same axes. Shade the region between the two curves and x = 1 and x = 3. c Find the exact area between f (x) and g(x) from x = 1 to x = 3 using a CAS calculator.
ThInk a
WrITe/draW a f (x) = 4 , g (x) = x
1
State the two functions.
2
Let f (x) = g(x) to find the values of x where the graphs intersect.
For points of intersection, x =
3
Solve for x.
x2 = 4 −4=0 (x − 2)(x + 2) = 0 x = −2 and x = 2
x
4 x
x2
b Sketch f (x) and g(x) on the same axes and
b
y
shade the region between the two curves from x = 1 to x = 3.
f(x) = x4– g(x) = x 01 2 3
c
1
State the area as the sum of two integrals for the two sub-intervals.
2
On the Main page, using the soft keyboard, tap: • ) • • P Complete the entry line as: 2 4 3 4 ∫ 1 ( x − x) dx + ∫ 2 ( x − x ) dx Press E.
3
Write the area with the correct units.
c Area =
∫1 ( x − x 2
4
x
) dx + ∫ ( x − 4x ) dx 3
2
4 ∴ The area is 4 log e + 1 square units. 3
Chapter 9
Integration
463
rememBer
1. If two curves f (x) and g(x) do not intersect over the interval [a, b] and f (x) > g(x), then the area enclosed by the two curves and the lines x = a and x = b is found by using the formula: b b b x ) − g( x)] x dx ∫ f ( x) dx − ∫ g( x) dx = ∫ [ f ( x) a
a
a
2. If two curves f (x) and g(x) intersect over the interval [a, b], it is necessary to find the points of intersection and hence find the area of each section, because sometimes f (x) > g(x) and sometimes g(x) > f (x). 3. If sketch graphs are not used to determine which is the upper curve, then it is necessary to take the absolute value or positive value of each integral. exerCISe
9h
areas between two curves 1 We32a State the definite integral that will find the shaded areas on each graph below. y y y a b c 2 y=x
y = 2x
y = 3x
y=x y=x+1
y
d
y = x2
e
y
2 x y = 8 − x2
0
−2
g
0
x
0 1
y
y=x 01
2
x
1
y = x3
0
h
y
f
y = 3x
x
3
x 1 y = 4 − x2
0
x
2
y
−1
y = ex
x 1 y = 9 − x2
0
y = x2 − 5
−1 0
x
1
y = −4 y=−e
x
2 We32b Find each of the areas in question 1.
y
f(x)
3 mC Which one of the following does not equal the shaded area? 5
5
∫1 g( x) dx + ∫5 f ( x) dx
f ( x ) dx − ∫ g( x ) dx
D
∫1 [g( x) − f ( x)] dx
C
∫1
E
∫5 [ f ( x) − g( x)] dx
5
1
g(x)
1
B
∫1 g( x) dx − ∫1 5
5
f ( x ) dx
A
0 1
5
1
5 y
g(x)
4 mC The area bound by the curves f (x), g(x) and the lines x = −3 and x = 1 at right is equal to: A
464
−3
∫− 1
[ f ( x ) − g( x )] dx
x
B
f(x)
−1
∫− 3 [ f ( x) + g( x)] dx
maths Quest 12 mathematical methods CaS for the Casio ClassPad
−4
−3 −1
0
x
C E
−1
∫− 3
[ g( x ) − f ( x )] dx
D
−3
−1
∫− 3 [ f ( x) − g( x)] dx
∫− 1 [ f ( x) + g( x)] dx y
5 MC The shaded area at right is equal to: A B
4
g(x)
∫0 [ f ( x) − g( x)] dx 3
f(x) 4
∫0 [g( x) − f ( x)] dx + ∫3 [ f ( x) − g( x)] dx
0
4
C
∫0 [g( x) − f ( x)] dx
D
∫0 [ f ( x) − g( x)] dx
E
∫0 [ f ( x) − g( x)] dx + ∫3 [g( x) − f ( x)]
3 4
x
3
3
4
6 WE33 In each of the following: i find the values of x where the functions intersect ii sketch the graphs on the same axes iii hence, find the area bound by the curves. a y = 4x and y = x2 b y = 2x and y = 3 − x2 2 2 c y = x − 1 and y = 1 − x d y = x2 − 4 and y = 4 − x2 2 2 e y = (x + 1) and y = 1 − x f y = x and y = x2 7 WE34 i Find the values of x where the functions intersect. ii Sketch the graphs on the same axes. iii Find the area between f (x) and g(x) giving an exact answer. a y = x3 and y = x b y = 3x2 and y = x3 + 2x 8 Find the area between the pairs of curves below, over the given interval. a y = x3, y = x2, x ∈ [−1, 1] b y = sin (x), y = cos (x), x ∈ [0, π] c y = (x − 1)2, y = (x + 1)2, x ∈ [−1, 1] d y = x3 − 5x, y = 6 − 2x2, x ∈ [0, 3] 1 e y = , y = 4x, x ∈ [ 14 , 1] x f y = e x, y = e−x, x ∈ [0, 1]
π g y = 2 cos (x), y = x − π , x ∈ [0, ] 2 2 x − x − h y = e , y = e , x ∈ [ 2, 1] 9 Find the area between the curve y = ex and the lines y = x, x = 1 and x = 3. 10 Find the area between the curve y = x2 and the lines y = x + 3, x = 1 and x = 3. 2 11 Calculate the area between the curves y = sin (2x) and y = cos (x) from x = 0 to x = π . 2 12 Calculate the area between the curves y = 3 − sin (2x) and y = sin (2x) from x = 0 to x = π . 4 13 Find the exact area bound by the curves y = ex and y = 3 − 2e−x. y 1 14 The graph at right shows the cross-section of a bricked archway. (All measurements are in metres.) a Find the x-intercepts of f (x). b Find the x-intercepts of g(x). c Find the cross-sectional area of the brickwork.
f(x) = 4 − –4 x2
x 0 g(x) = 3 − 1–3 x2
Chapter 9 Integration
465
15 The diagram below shows the outline of a window frame. If all measurements are in metres, what is the area of glass which fits into the frame? y
y = 2 1–2 − 2x2 y = 2x2
− 1–2
0
x
1– 2
y
16 The diagram at right shows the side view of a concrete bridge. (All measurements are in metres.) Find: a the x-intercepts of the curve b the length of the bridge c the area of the side of the bridge d the volume of concrete used to build the bridge if the bridge is 9 metres wide. 17 The cross-section of a road tunnel entrance is shown at right. (All measurements are in metres.) The shaded area is to be concreted. Find: a the exact area, above the entrance, which is to be concreted b the exact volume of concrete required to build this tunnel if it is 200 metres long.
9I
5 2
x y = 4 − —– 100
0
x
y
0
x f(x) = 5 sin (π–– 30 )
x
Average value of a function Consider the function y = f (x). The average value, yav, for the function y = f (x) over the interval [a, b] is given by: b 1 yav = f ( x ) dx. b − a ∫a b
This can be rearranged to give yav × (b − a) = ∫ f ( x ) dx. a
y
y = f(x)
yav a
x
b
Geometrically, the average value of the function is the height, yav, of the rectangle of width (b − a) that has the same area as the area under the graph of y = f (x) for the interval [a, b]. Worked Example 35
Find the average value of f (x) = 2x2 for the interval [1, 4]. Find the value of x that corresponds to the average value. Think 1
466
Write the relationship for the average value.
Write
yav =
b 1 f ( x ) dx b − a ∫a
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
4 1 2 x 2 dx ∫ 1 4 −1
2
Identify f (x) = 2x2 and substitute in the values for a and b.
yav =
3
Antidifferentiate.
1 2x3 yav = 3 3 1
4
Evaluate the definite integral and multiply by 1 . 3
4
1 2 × 43 2 × 13 yav = − 3 3 3
=
1 128 2 × − 3 3 3
1 126 × 3 3 = 14 ∴ The average value is 14.
5
Solve f (x) = 14 to find x.
=
2 x2 = 14 x2 = 7 x=± 7
6
∴x = 7 as x ∈ [1, 4].
Choose the value of x that satisfies the given interval.
Worked Example 36
Find the average value of f (x) = loge (2x) for the interval [2, 4]. Give your answer in exact form. Think 1
Write the relationship for the average value and substitute the values for a and b.
2
The integral of loge(2x) is not covered in this course. The average value can only be evaluated using a CAS calculator. On the Main page, using the soft keyboard, tap: • ) • • P Complete the entry line as: 1 4 (ln (2 x )) dx 2 ∫2 Press E.
3
Write the answer.
Write
yav =
4 1 log e (2 x ) dx ∫ 2 4−2
∴ The average value f (x) = loge (2x) for the interval [2, 4] is 4 loge (2) − 1.
Chapter 9 Integration
467
rememBer
The average value, yav, for the function y = f (x) over the interval [a, b] is given by yav =
b 1 f ( x ) dx . ∫ b−a a
exerCISe
9I
average value of a function 1
2
3
4
We35
a y = x3 − x, x ∈ [1, 3]
π b y = sin (x), x ∈ 0, 6
c y = x, x ∈ [1, 4]
d y = e3x, x ∈ [0, 2]
We36 Find the average value of the function, in exact form, for the given interval.
a y = x loge (x), x ∈ [2, 5]
π b y = tan (2x), x ∈ 0, 6
c y = xex, x ∈ [0, 4]
d y = x x + 1, x ∈ [1, 7]
− We35 For the function y = e 2x for x ∈ [−1, 0], find:
a the average value of the function b the corresponding x-value in exact form.
mC The average value of the function f (x) = loge (2x + 1) over the interval [0, 4] is:
A 5
log e (9) 4
B loge (9)
C
1 [ 9 log e (3) − 4 ] 4
D 9 loge (3) − 4
mC The average value of the function y = sin (2x) over the interval 0,
A
9J
Find the average value, in exact form, of the function for the given interval.
4 π
B 2π
C
π 8
D
1 2
E
2 π
E
−9
π is: 4
log e (3) + 4 4
further applications of integration Differentiation can be applied to rates of change and related rates. If the rate of change of a function is known, antidifferentiation allows the original function to be found. So integration has many practical applications.
Worked examPle 37
eBook plus
The rate of change of position, velocity, of a particle travelling in a straight line is dx − given by = 40 − 10 e 0.4 t m/s, t ≥ 0, where x is measured in metres and t in dt seconds. a Find the velocity: i initially ii after 10 seconds, correct to 2 decimal places. b Find the time taken to reach a velocity of 35 m/s. dx c Sketch the graph of against t. dt d Find the total distance travelled by the particle in the first 10 seconds. 468
maths Quest 12 mathematical methods CaS for the Casio ClassPad
Tutorial
int-0569 Worked example 37
Think a
i
write 1
The initial velocity occurs when t = 0. Substitute t = 0 into
2
ii
1
dx . dt
1
Answer with the correct units. dx Substitute = 35. dt
dx − = 40 − 10 e 0.4 t dt − = 40 − 10 e 0.4 × 0 = 40 − 10 e 0 = 40 − 10 = 30 ∴ Velocity is initially 30 m/s. dx − = 40 − 10 e 0.4 ×10 dt − = 40 − 10 e 4 = 39.82
Answer with the correct units. dx Substitute t = 10 into . dt
2
b
a
∴ After 10 seconds, the velocity is 39.82 m/s. dx − b = 40 − 10 e 0.4 t dt 35 = 40 − 10 e
2
Solve for t.
−5
= − 10 e
− 0.4 t
− 0.4 t
e −0.4 t = 0.5 − 0.4t = ln (0.5) ln (0.5) − 0.4 = 1.73 s
t=
3
c
d
1
∴ Time taken is 1.73 seconds.
Answer the question correct to 2 decimal places. dx = 40 − 10 e −0.4x , dt open the Graph & Tab page and complete the function entry line as: y1 = 40 - 10e 0.4x Tick the y1 box and tap !. Note the viewing window settings: XMin: 0, XMax: 15, YMin: 0 and YMax: 45 To graph
2
When drawing your graph, label the axes with the given variables. Note the horizontal asymptote dx = 40 is not displayed. The dt asymptotic behaviour of the function, however, is clearly visible.
1
Distance travelled = area under the graph State the distance as a definite integral.
c
d dx = 40 − 10 e −0.4 t
dt
x=
10
∫0
(40 − 10 e
−0.4 t )dt
Chapter 9 Integration
469
2
Antidifferntiate.
3
Evaluate the integral.
4
State the distance travelled with correct units.
10
− = 40t + 25e 0.4 t 0 −
= (40 × 10 + 25e 0.4 ×10 ) − (40 × 0 + 25e − = (400 + 25e 4 ) − (25e 0 ) −4 = 400 + 25e − 25 − = 375 + 25e 4 = 375.46
− 0.4 × 0
∴ The distance travelled in the first 10 seconds is 375.46 metres.
Worked Example 38
The rate of change of pressure, P atmospheres, of a given mass of gas with respect to its volume, dP − k , k > 0. = dV V 2 a If d P = −5 when V = 10, find k. dV b Find the pressure, P, as a function of V given that when P = 10 atmospheres, V = 50 cm3. c Find the volume when the pressure is 20 atmospheres. V cm3, is given by
Think a
1
Write
dP − Substitute the conditions = 5 and V = 10 dV into the given rate.
a
dP − k = dV V 2 −5
b
2
Simplify.
3
Solve for k.
1
Write the rate with the value of k found previously.
2
Antidifferentiate.
=
−
k
(10)
2
−
k 100 ∴ k = 500 b dP = − 500 2 dV V −5
=
P=∫
−
500 V
2
dV
= − 500 ∫ V
470
Substitute the given conditions P = 10 and V = 50 to find c.
4
Write the relationship for P.
−2
500 +c V 500 P= +c V 500 10 = +c 50 c=0 500 ∴P= V =
3
)
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
dV
c
1
Substitute P = 20.
2
Solve for V.
3
State the answer with correct units.
Exercise
9J
c
500 V 500 20 = V 500 V= 20 ∴ The volume is 25 cm3. P=
Further applications of integration 1 If f ′(x) = (2 − x)2 and the y-intercept of f (x) is 4 , find the rule for f (x). 3 π dy 2 If = 1 − 4 cos (2 x ) and the y-intercept is 2, find the exact value of y when x = . 12 dx 3 The rate of deflection from a horizontal position of a 3-metre diving board when an 80-kg dy − person is x metres from its fixed end is given by = 0.03( x + 1)2 + 0.03, where y is the dx deflection in metres. y (Metres) 0 Board
(Metres) x Deflection
a What is the deflection when x = 0? b Determine the equation that measures the deflection. c Hence, find the maximum deflection. 4 On any day the cost per item for a machine producing n items is given by dC = 40 − 2e 0.01n, where n ∈ [0, 200] and C is the cost in dollars. dn a Use the rate to find the cost of producing the 100th item. b Express C as a function of n. c What is the total cost of producing the first 100 items? d Find the average cost of production for: i the first 100 items ii the second 100 items. 5 WE37 The rate of change of position (velocity) of a racing car travelling down a straight dx stretch of road is given by = t (16 − t ), where x is measured in metres and t in seconds. dt a Find the velocity when: i t = 0 ii t = 4. b Determine: i when the maximum velocity occurs ii the maximum velocity. dx c Sketch the graph of against t for 0 ≤ t ≤ 16. dt
Chapter 9 Integration
471
d Find the area under the graph between t = 0 and t = 10. e What does this area represent?
6 WE38 The rate at which water is pumped out of a dam, in L/min, t minutes after the pump is started dV πt = 5 + cos . started is 40 dt a How much water is pumped out in the 40th minute? b Find the volume of water pumped out at any time, t, after the pump is started. c How much water is pumped out after 40 minutes? d Find the average rate at which water is pumped in the first hour. e How long would it take to fill a tank holding 1600 litres? 7 The rate of flow of water into a hot water system during a 12-hour period is thought to be dV πt = 10 + cos , where V is in litres and t is the number of hours after 8 am. 2 dt dV a Sketch the graph of against t. dt b Find the length of time for which the rate is above 10.5 L/h. c Find the volume of water that has flowed into the system between: i 8 am and 2 pm ii 3 pm and 8 pm. 8 The roof of a stadium has the shape given by the function f : [−25, 25] → R, f (x) = 20 − 0.024x2. The stadium is 75 metres long and its cross-section is shown at right. a Find the volume of the stadium. b The stadium is to have several airconditioners strategically placed around it. Each can service a volume of 11 250 m3. How many airconditioners are required?
y (metres) 20 5 −25
(metres) x 25
0
y
y = x3
9 The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full. 10 For any point P on the curve y = x3, prove that the area under the curve is one quarter of the area of the rectangle.
472
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
P
0
x
11 The arch of a concrete bridge has the shape of a parabola. It is 6 metres high and 8 metres long. a Find the rule for the function corresponding to the arch of the bridge. b Find the area of the shaded region. c If the bridge is 10 metres wide, find the volume of concrete in the bridge. 12 In the figure at right f (x) and g(x) intersect at O and B. a Show that the coordinate of B is (loge (2), 1). b Find the exact area of the region bound by f (x) and g(x). c Show that the sum of the areas under f (x) and g(x), from x = 0 to x = loge (2), is equal to the area of the rectangle OABC.
y (metres) 7 6
−4 y
0
4 5
(metres) x
f(x) = ex − 1 B g(x) = −2e−x + 2
C
O
A
x
Chapter 9 Integration
473
Summary Antidifferentiation rules
• The relationships between f (x) and ∫ f ( x ) dx are: f (x)
∫ f ( x) dx
a
ax + c
ax n
ax n + 1 +c n +1
(ax + b)n
(ax + b) n + 1 +c a( n + 1)
1 x
loge |x| + c
1 ax + b
1 log |ax + b| + c a
ex
ex + c
ekx
1 kx e +c k −
sin (ax) cos (ax) • ∫ [ f ( x ) ± g( x )] dx =
∫ f ( x) dx ± ∫ g( x) dx
∫ g( x) dx = f ( x) + c, where g(x) = f ′(x) dx
•
1 cos (ax) + c a 1 sin (ax) + c a
∫ kf ( x) dx = k ∫ f ( x) dx • ∫ f ( x) dx is the indefinite integral •
Definite integrals
• The fundamental theorem of integral calculus: b
∫a
b
f ( x ) dx = [ F ( x )]ba = F (b) − F (a) where F (x) is an antiderivative of f (x).
•
∫a
•
∫a f ( x) dx = ∫a f ( x) dx + ∫c
•
∫a [ f ( x) ± g( x)] dx = ∫a
f ( x ) dx is the definite integral
b
c
b
b
b
∫a
•
b
f ( x ) dx =
−
a
∫b
•
b
b
∫a kf ( x) dx = k ∫a
f ( x ) dx
f ( x ) dx, a < c < b b
f ( x ) dx ± ∫ g( x ) dx a
f ( x ) dx
Graphs of the antiderivative function
• • • •
474
For a polynomial function, the graph of f (x) is one degree higher than the graph of f ′(x). The x-intercepts of f ′(x) are the x-coordinates of the turning points of f (x). When f ′(x) is above the x-axis, the gradient of f (x) is positive. When f ′(x) is below the x-axis, the gradient of f (x) is negative.
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
Approximating areas under curves
• An approximation to the area between a curve and the x-axis can be found by dividing the area into a series of rectangles that are all the same width. The approximation is found by finding the sum of all the areas of the rectangles. • For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. • For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation. Area under curves b
• Area = ∫ f ( x ) dx, if f ( x ) > 0 for x ∈[a, b] a
y
•
b
Area = − ∫ f ( x ) dx, if f ( x ) < 0, or a
b
∫ a f ( x)dx , for x ∈ [a, b] y
y = f(x)
y = f(x) a
b x
0
0
• Area =
b
∫c
a
x y
c
f ( x ) dx − ∫ f ( x ) dx
y = f(x)
a
b
b
= ∫ f ( x ) dx + c
c
∫a f ( x) dx , if
f ( x ) > 0 for x ∈[c, b]
and f (x) < 0 for x ∈[a, c]
A1 0 A 2
a
c
x
b
Area between curves y
b
• Area = ∫ [ f ( x ) − g( x )] dx, if f ( x ) > g( x ) for x ∈[a, b]
f(x)
a
g(x) 0 a
c
b
a
c
• Area = ∫ [ g( x ) − f ( x )] dx + ∫ [ f ( x ) − g( x )] dx,
b
y
x
g(x) f(x)
if g(x) > f (x) for x ∈ [a, c] and f (x) > g(x) for x ∈[c, b] a 0
c
b
x
Limiting value of a sum
•
b
∫a
0
∑ f ( xi ) δ xi δ x→0
f ( x )dx = lim
i =1
Average value of a function
• yav =
1 b−a
b
∫a
f ( x )dx
Chapter 9 Integration
475
chapter review x
Short answer
1 Find the equation of the curve f (x) if it passes 3x3 − 2 x 2 through (1, −3) and f ′( x ) = . x dy πx = cos + k, where 2 A particular curve has 4 dx k is a constant, and it has a stationary point (2, 1). Find: a the value of k b the equation of the curve c the value of y when x = 6. dy 3 If y = sin (x2 + 2x), find and hence dx antidifferentiate (x + 1) cos (x2 + 2x).
12 The graph of f: R → R, f ( x ) = e 2 + 1 is shown. The normal to the graph of f where it crosses the y-axis is also shown. y
0
a Find the equation of the normal to the graph of f where it crosses the y-axis. b Find the exact area of the shaded region.
4 A curve has a gradient function f ′(x) = 2ex + k. It has a stationary point at (0, 3). Find: a the value of k b the equation of the curve f (x).
Exam tip a Students often had difficulty correctly finding the y-intercept; (0, 1) was a common answer, leading to the normal being y = −2x + 1. Quite a few students differentiated but then did not find the value of the derivative at x = 0 and simply substituted the expression for the derivative into the equation of the normal, which led to incorrect attempts at algebraic manipulation and nonlinear normals. Some students found a tangent instead. b Most students correctly set up an area expression but some were unable to proceed due to their ‘normal’ not being a linear expression. Many had the correct terminals, although 2 instead of 1 was a popular incorrect value. Arithmetic errors appeared frequently and subtraction mistakes in the integrand were also common. Some students also lost ‘1’ from the equation of the curve. The use of ‘dx’ was surprisingly good. A few students used the area under the curve minus the triangle with some success.
5 Calculate the exact area between the curve y = ex − 4 and the lines y = x, x = 1 and x = 2. (Hint: y = ex − 4 and y = x do not intersect between x = 1 and x = 2.) 6 Evaluate each of the following definite integrals. 0 − 9 a ∫ − dx 1 (2 x + 3) 4 b
∫
2π 3 π 3
cos (2 x ) dx
7 Given that
k
∫0 (4 x − 5) dx = − 2, find two
possible values for k. 1 . x−2 b Find the exact area between the graph of f (x), the x-axis and the lines x = 3 and x = 6.
8 a Sketch the graph of the function f ( x ) =
9 Find the area bound by the x-axis and the curve g(x) = (4 − x)(6 + x). 10 Calculate the area between the curve y = 2 cos (x) π and the lines y = −x, x = 0 and x = . 2 11 Use the method of left rectangles to approximate the area under the curve y = x2 + 1, from x = 1 to x = 4, using interval widths of 1 unit. 476
x
[Assessment report 1 2007]
[© VCAA 2007]
Multiple choice
1 The antiderivative of 4 x 3 − A x4 − loge (1 − x) + c B x4 + loge (1 − x) + c c 16x4 − loge (1 − x) + c D 16x4 + loge (1 − x) + c 1 +c E x 4 − (1 − x )2
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
1 is: 1− x
2 The indefinite integral ∫ (5 x − 4)4 dx is equal to: A 25(5x − 4)5 + c
B 5(5x − 4)5 + c
C (5x − 4)5 + c
D
E
1 (5 x − 4) 5 25
1 (5 x − 4) 5 5
+c
−
−3
−2 9
E
−8(3x
(3x + 4)
D
−3 −3x
−1 2
−3x
e
−5
(3x + 4)
+ 4)
4 The antiderivative of 6e − A −2e 3x + c − C −18e 3x + c E
−2 9
is: − B −3e 3x + c − D −2e 3x +1 + c
+c
x 5 The indefinite integral ∫ cos − 3 sin ( 3 x ) dx 3 is equal to: x A sin + cos (3x) + c 3
() x B sin ( ) + 3 cos (3x) + c 3 x C 3 sin ( ) + cos (3x) + c 3 x D 3 sin ( ) + cos (3x) + c 3 x E 3 sin ( ) − cos (3x) + c 3 1 3
−
6 An antiderivative of x3 + sin (4x) + e4x is: A 4[x4 − cos (4x) + e4x] x 4 + cos (4 x ) + 14 e 4 x
B
1 4
C
1 4 x 4
D
1 4
[x4 − cos (x) + e4x]
E
1 4
[x4
− 4 cos (4x) + 14 e4x
A
3 3e x + 3 x
2
+3
+c
3
+3x
+ 1)e x + 3 x
is:
is 3( x 2 + 1)e x (x2 B D
3
1 3( x 2 + 1) 1 x3 +3 x e 3
,
+c
+c
E undefined
− x 2 10 If the derivative of loge (5 − x2) is then the 5 − x2 x antiderivative of is: 5 − x2 A − 12 loge (5 − x2) + c B −2 loge (5 − x2) + c
C 12 loge (5 − x2) + c D 2 loge (5 − x2) + c E undefined 11 The approximation for the y (3.5, 9) area under the graph at (3, 6) right from x = 2 to x = 4, (2.5, 4) using the ‘lower (2, 3) rectangles’ is: A 22 sq. units B 14 sq. units x 0 3 2 4 C 11 sq. units D 10 sq. units E 20 sq. units 12 The area under the graph y at right from x = −5 to (−5, 8) x = −1 can be (−4, 6) approximated by the area (−3, 5) (−2, 4) of the ‘upper rectangles’ and is equal to: A 20 sq. units x −5 −1 0 B 21 sq. units C 23 sq. units D 11 12 sq. units E 10 sq. units 13 The interval [0, 4] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 0 = x0 < x1 < . . . < xn − 1 < xn = 4. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n
∑ ( xiδ x) is equal to: δ x→0
Then lim
7 If f (x) has a stationary point at (0, 3) and f ′(x) = ex + k, where k is a constant, then f (x) is: A ex − 2x + 2 B ex − x + 2 C ex − x + 2 −x x D e +x+2 E e + 2x + 1
0
8 If the derivative of (x − x2)8 is 8(1 − 2x)(x − x2)7 then an antiderivative of 24(1 − 2x)(x − x2)7 is: A 2(x − x2)8 B 3(x − x2)8 C 12 (x − x2)8 1 2 8 2 8 D 3 (x − x ) E 8(x − x )
+3x
+c
e4x]
− cos (4x) +
3
then the antiderivative of
C e3 x
+c
3 An antiderivative of 2(3x + 4) 4 is: − − A − 23 (3x + 4) 3 B − 23 (3x + 4) 3 + 5 C
9 If the derivative of e x
i =1
A
∫ 4 x dx
B
4
∫0
x2 dx 2
C 0
D 4 E 8
exam TIP The definition of the definite integral as the limiting value of a sum, and its corresponding representation in integral form, is a fundamental property of integral calculus and must be understood.
[Assessment report 1 2003]
[© VCAA 2003]
Chapter 9
Integration
477
Questions 20 to 22 apply to the curve with equation f (x) = ex − 1.
4
14 The expression ∫ (3 x − x ) dx is equal to: 0
A 2 B 8 C −2 12 D 20 E 16
20 The graph of f (x) is best represented by: y y A B f(x) f(x) 2
15 The exact value of the definite integral 2
C
−4
A −e − B 2e4 − e 4 − 4 C e − 2e 4 −4 4 D e + e − E e4 − e 4 3e4
()
f(x)
1 8 The shaded area on the graph at right is: A 20 sq. units B −20 sq. units C −16 sq. units D 16 sq. units E −18 sq. units 19 The area bound by the curve on the graph at right and the x-axis is equal to: 5 20 12 1
5 1012
sq. units
5
D −10 12 sq. units 7
E 20 12 sq. units
y
D
−1
f(x) 0
x
f(x) x
−1
21 The area bound by the graph of f (x), the x-axis and the line x = 2 is equal to: A e2 − 1 B e2 − 2 C e2 + 1 D e2 + 2 E e2 − 3
y
y = (x − 2)3
0
4 x
2
22 The area bound by the graph of f (x), the y-axis and the line y = e2 − 1 is equal to: A e2 − 5 B e2 − 3 C e2 + 2 D e2 + 1 E 5 − e2 Use the graph below to answer questions 23 and 24. y
y 0
−2
2
x
y = 2x + 3 0
−1 y = −1 − 3x2
y
y = x2
x
y = x(x + 2)(x − 3)
2 3 The two graphs intersect where x is equal to: A 1 and −3 B −1 and 3 C 1 and 2 D −1 and − 2 E 1 and 3 24 The area bound by the two graphs is equal to: 2 1 A 10 3 sq. units B 7 3 sq. units 1
C −7 3 sq. units
sq. units
B 21 12 sq. units
x
0
x
−1 0
E 3 2 1 7 The shaded area on the graph at right is equal to: A 12 sq. units B 16 sq. units C 10 sq. units D 4 sq. units E 8 sq. units
−1 y
E
D 3 3
478
y
0
π x 16 The exact value of ∫ − 2 cos dx is: 0 3 A −3 B 3 C −3 3
C
x
−
∫ − 2 (4e2 x − 2e 2 x ) dx is:
A
0
1
D 11 3 sq. units
2
0
x
E 6 3 sq. units 25 The average value of the function y = cos (2x) over π the interval 0, is: 4 4 2 π A B c π π 8 1 d E 2π 2
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
Extended response
1 From past records it has been found that the cost rate of maintaining a certain car is dC = 75t 2 + 50t + 800, where C is the accumulated cost in dollars and t is the time in years since the car dt was first used. Find: a the initial maintenance cost b C as a function of t c the total maintenance cost during the first 5 years of use of the car d the total maintenance cost from 3 to 5 years e the maintenance cost for the second year. 2 Over a 24-hour period on a particular autumn day, starting at 12 midnight, the rate of change of the temperature for Melbourne was approximately dT − 5π πt cos , where T is the temperature = 12 dt 12 in °C and t is the number of hours since midnight when the temperature was 10 °C. Find: a the temperature at any time, t b whether the temperature reaches 17 °C at any time during the day c the maximum temperature and the time at which it occurs d the minimum temperature and the time at which it occurs e the temperature at i 2 am i i 3 pm f the time when the temperature first reaches 14.33 °C.
x
3 The diagram at right shows part of the curve with equation y = e 2 . Find: a the coordinate of point A b the equation of the normal to the curve at point A c the coordinate of point B d the coordinate of point C e the area bound by the curve and the lines AB and BC. 4 a Find the derivative of x loge (x). b Hence, find an antiderivative of loge (x). The cross-section of a platform is shown at right. (All measurements are in metres.) c Find the height of the platform. d Find the cross-sectional area of the platform. e Find the volume of concrete required to build this platform if it is 20 metres long.
y –x
Normal
y = e2
B A
C 0
2
x
y 1 0
e−2
e x 1 f(x) = logex
Chapter 9 Integration
479
5 A thick metal pipe is filled with boiling water and is kept boiling. The temperature, T °C, of the metal in the pipe decreases relative to its distance, x cm, from the centre of the pipe. dT − 20 It is known that and 4 ≤ x ≤ 8. = dx x a Find the rate of change of the temperature in the metal on the outside of the pipe. b Express T as a function of x. c Find the temperature of the metal, correct to 2 decimal places: i when x = 6 cm ii on the outside of the pipe. eBook plus Digital doc
Test Yourself Chapter 9
480
maths Quest 12 mathematical methods CaS for the Casio ClassPad
x
eBook plus
Activities
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on integration. (page 416) 9A Antidifferentiation Digital doc
• SkillSHEET 9.1: Practise substitution and evaluation. (page 425) 9c Integration by recognition Tutorial
• WE 13 int-0564: Watch a worked example on performing integration by recognition. (page 430)
9H Areas between two curves
Digital doc
Tutorial
• WorkSHEET 9.1: Determine functions using antidifferentiation. (page 435) 9D Approximating areas enclosed by functions Interactivity int-0254
• Approximating areas enclosed by functions: Consolidate your understanding of varying approximations to areas enclosed by functions. (page 435) Tutorial
• WE 18 int-0565: Watch a worked example on the approximation of the area under a curve. (page 435) 9E The fundamental theorem of integral calculus Tutorial
• WE 21 int-0566: Watch a worked example on evaluating definite integrals. (page 442)
int-0568: Watch a worked example on • WE 34 int-0801: calculating the area between two curves using CAS. (page 463)
9J Further applications of integration Tutorial
• WE 37 int-0569: Watch a worked example on applications of integration. (page 468) Chapter review Digital doc
• Test Yourself: Take the end-of-chapter test to test your progress. (page 480) To access eBookPLUS activities, log on to www.jacplus.com.au
Digital docs
• Definite integrals: Investigate the properties of definite integrals. (page 442) • SkillSHEET 9.2: Practise subtracting function values. (page 446) 9F Signed areas Tutorial
• WE 28 int-0567: Watch a worked example on finding the area bound by an exponential curve above and below the x-axis. (page 450) Digital doc
• WorkSHEET 9.2: Approximate areas between curves and calculate areas between curves using integrals. (page 454)
Chapter 9 Integration
5_61_60255_MQ12MM_CAS_09.indd 481
481
9/11/09 9:15:27 AM
EXAM PRACTICE 3
Short answer
Chapters 1 TO 9
y
40 minutes
4
1 The graph of f (x): R → R, where f (x) = 10xe-x is graphed below.
3
y 4
2 1
2 −2
0
2
4
6
−2
a By selecting approximate values from the graph, determine a left-rectangle approximation to the area bounded by the graph, the x-axis and the lines x = 1 and x = 5 by dividing the area into four strips of equal width. dy b If y = (x + 1)e−x, find . dx c Hence determine an exact value for the shaded area. 3 marks
2 Write down an antiderivative of: a y = 4x3 - 2 cos (x) + e x 3 b y = x + x
3
x
4 marks
Multiple choice
8 minutes
Each question is worth 1 mark. 1 Which integral could be evaluated in order to calculate the area enclosed between the graphs of y = f (x) and y = g(x) and the vertical lines x = a and x = b? f x=b
4 The area of the region bound by the curve with x equation y = a cos , the x-axis and the line 3 2 marks
5 A spa is being filled with water at the rate of 30 litres per minute. The volume (V litres) of water in the spa is related to its depth (x cm) by V = 250 loge (x). At what rate (cm per min) is the depth of the spa increasing? 2 marks
6 The graph of f : R → R, f (x) = (x - 1)(5 - x) is shown. The tangent to the graph of f where it crosses the x-axis is also shown.
482
2
a Find the equation of the tangent to the graph of f where it crosses the x-axis. b Find the exact area of the shaded region.
x=a
2 marks
with equation x = π is 3. Find a.
1
y
3 A is (0, 2), a point on the y-axis. P is a point on the parabola y = x2 such that the length of AP, the line segment from the A to P is a minimum. a Find the coordinates of P. b Find the minimum length of AP. 2 marks
0
−1
8x
g x
a
A ∫ ( f ( x) − g( x )) dx b b
C ∫ ( f ( x) − g( x )) dx a
b
B ∫ ( f ( x) + g( x )) dx a a
D ∫ ( f ( x) + g( x )) dx b
a
E
∫ (g( x) − f ( x)) dx b
π 2 The average value of the function y = 3 sin (2 x + ) 3 π over the interval [0, ] is: 3 3 3 A 0 B 4 1 3 3 C 2 D 4 2 E
27 4π
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
2
3 Using the fact that ∫ f ( x ) dx = 5, what is the value 1
1
of ∫ (3 f ( x ) − 1) dx? 2 −16
B −14 E 16
A D 14
C −12
4 A function with the rule y = f (x) is shown in the first sketch below. y
y
x
x
Which of these rules could correspond to the second sketch? A y = f ′ (x) B y = f −1(x) C y = −f (x) D y = f (−x) 1 E y= f ( x) 5 If f (x) = (2x + 1)2 and f (0) = 1 then the antiderivative of f (x) will be equal to: (2 x + 1)3 2 + 3 3 1 C ((2 x + 1)3 + 5) 6 E 4x3 + 4x2 + x + 1
B 4x + 1
A
D 2(2x + 1) + 1
EXTENDED RESPONSE
30 minutes
The volume of a cylindrical soft drink can is 250 mL. The volume of a cylinder can be calculated using V = π r2h, the area of a circle is given by A = π r2 and its perimeter is given by P = 2π r. Also 1 mL = 1 cm3. a Create an expression for the total surface area of the can, S, in terms of r and h.
r
h
1 mark
b Use the fact that the volume is 250 mL to show that S can be expressed in terms of r 500 only as S = 2π r 2 + . 1 mark r 500 y c The graphs at right are of y = 2π r2 and y = for 0 ≤ r ≤ 10. Use these graphs r 700 600 500 to create a sketch of S = 2π r 2 + on the same axes. 500 r 1 mark d Hence obtain an estimate, to 1 decimal place, of the radius corresponding to the minimum total surface area.
1 mark
400 300 200
100
e Using calculus determine an exact value for the radius corresponding to the 0 2 minimum surface area. f i Write down a decimal approximation to 2 decimal places for the radius corresponding to the minimum total surface area. ii Confirm using the sign of gradient test that the stationary point located is in fact a minimum. iii Write down the value of the minimum surface area to the nearest cm2.
4
6
8 10 x
1 + 1 = 2 marks
g If the cost of the material for the bottom and top of the can is twice the cost for the sides determine the radius of the can for minimum total cost. h The actual radius for a can of soft drink is 3.0 cm. Determine to 2 decimal places the ratio cost of material for top and bottom which would be consistent with this can having the minimum cost of material for the sides cost of materials.
2 marks
2 marks
eBook plus Digital doc
Solutions Exam practice 3
Exam practice 3
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