Question 1: Find the volume of the solid formed by revolving the region bounded by the graphs of y = sec(x) tan(x), y = 0, x = 0, and x =
π 3
about the line y = −1. (a) 2π
(b)
√ 3π
(c)
π2 3
(d)
2+
√ 3 π
(e)*
2+
π π 3
(f)
Solution: Washer method: inner radius r(x) = 1, outer radius R(x) = sec(x) tan(x) + 1.
Z
π/3
(sec(x) tan(x) + 1)2 − 1 dx
V =π 0
Z =π
π/3
sec2 (x) tan2 (x) + 2 sec(x) tan(x) dx
0
π/3 tan3 (x) =π + 2 sec(x) 3 h√ i0 √ = π ( 3 + 4) − (0 + 2) = (2 + 3)π
√
3+
π π 3
Question 2: 1 from x = 12x 23 21 (c)* (d) 24 12
Find the arc-length of the curve y = x3 + (a) 0
(b)
11 12
1 2
to x = 1. (e)
23 12
(f)
47 24
Solution: y 0 = 3x2 −
1 12x2
Since y 0 = a − b with ab = 1/4, we can use the trick compute this directly. p 1 + (y 0 )2 = 3x2 +
p 1 + (y 0 )2 = a + b. Alternatively,
1 12x2
Hence the arc-length is Z
1
L=
p 1 + (y 0 )2 dx
1/2 1
Z =
3x2 +
1/2
1 dx 12x2 1
1 12x 1/2 1 1 23 1 − − = = 1− 12 8 6 24 = x3 −
Question 3: Evaluate the integral below. Z 0
(b) 2π −
(a) 2π
1 3
(c) 2π −
2 3
2
x4 dx 4 + x2 (d) 2π −
4 3
(e) 2π −
8 3
(f)* 2π −
16 3
Solution 1: Since there is a factor of the form a2 + x2 , we may use the trigonometric substitution x = a tan(θ). Here let x = 2 tan(θ). Note that dx = 2 sec2 (θ)dθ and 4 + x2 = 4 sec2 (θ). Hence Z 0
2
x4 dx = 4 + x2
Z
x=2
x=0
16 tan4 (θ) · 2 sec2 (θ) dθ = 4 sec2 (θ)
Z
x=2
8 tan4 (θ) dθ
x=0
Now we use the fact that tan2 (θ) = sec2 (θ) − 1. Z Z 4 8 tan (θ)dθ = 8 tan2 (θ) sec2 (θ) − 1 dθ Z Z 2 2 = 8 tan (θ) sec (θ)dθ − 8 sec2 (θ) − 1dθ =
8 tan3 (θ) − 8 tan(θ) + 8θ 3
At this point we must either change the bounds from x to θ or the function from θ to x. Changing bounds: If 0 = 2 tan(θ), then θ = 0. If 2 = 2 tan(θ), then tan(θ) = 1 and θ = π/4. Hence π/4 Z 2 8 16 x4 3 dx = tan (θ) − 8 tan(θ) + 8θ = 2π − . 2 3 3 0 4+x 0 Changing the function: If x = 2 tan(θ), then tan(θ) = x/2 and θ = arctan(x/2). Hence 3 2 Z 2 x4 x 16 dx = − 4x + 8 arctan(x/2) = 2π − . 2 3 3 0 4+x 0 (Sneaky) Solution 2: By long division,
x4 4+x2
Z 0
2
= x2 − 4 + x4 dx = 4 + x2
16 . 4+x2
Z 0
Hence
2
x2 − 4 +
16 dx 4 + x2
2 x3 16 = − 4x + 8 arctan(x/2) = 2π − 3 3 0
Question 4: Compute the partial fraction expansion of
(x + 1)3 . x(x2 + 1)
One of the fractions has denominator x2 + 1. What is that fraction’s numerator? (a)* 2x + 2
(b) 3x + 2
(c) 4x + 2
(d) x + 3
(e) 2x + 3
(f) 3x + 3
Solution: The integrand is a fraction with numerator and denominator both polynomials, so we can try to use PFE. Since the degree of the numerator and denominator are equal, there will be a constant as well as fractions for each factor in the denominator. B Cx + D (x + 1)3 =A+ + 2 2 x(x + 1) x x +1 (x + 1)3 = Ax(x2 + 1) + B(x2 + 1) + (Cx + D)x Comparing coefficients of x3 and x0 gives A = 1 and B = 1. Now either expand the left side to x3 + 3x2 + 3x + 1 and compare coefficients of x2 and x or substitute in any two non-zero values of x to solve for C and D. (Substituting x = 0 is the same as looking at the coefficient of x0 , which we already did.) For example, (x = −1) (x = 1)
0 = −2 + 2 + C − D 8=2+2+C +D
So C = D and C + D = 4. This implies C = D = 2.
Question 5: For a certain positive real number C, the function ( 1 if 1 ≤ X ≤ C f (X) = 2x 0 otherwise. is a probability density function for a continuous random variable X. Find the mean of X. (a)
e 2
(b)
e−1 2
(c)
e−4 4
(d)
e2 2
(e)*
e2 − 1 2
Solution: Since f (X), is a PDF, Z
∞
C
Z f (x) dx =
1= −∞
1
1 1 1 dx = ln(x) = ln(C). 2x 2 2
Therefore 2 = ln(C) and so C = e2 . Furthermore, the mean µ is Z
∞
Z xf (x) dx =
−∞
1
e2
e2 1 e2 1 1 = dx = x − . 2 2 1 2 2
(f)
e2 − 4 4
Question 6: Determine whether the sequence below converges or diverges and find its limit. s 2n an = n + sin(n) √
(a) converges, lim an = 0
(b)* converges, lim an =
(d) diverges, lim an = +∞
(e) diverges, lim an = −∞
n→∞
n→∞
n→∞
2
n→∞
(c) converges, lim an = 2 n→∞
(f) diverges, lim an does not exist n→∞
Solution 1: Use a combination of the sandwich lemma and the continuous function theorem: −1 ≤ sin(n) ≤ 1 2n 2n 2n ≤ ≤ n+1 n + sin(n) n−1 The limits of the left and right sequences are both 2, so the middle sequence also goes to 2. s √ 2n lim an = lim = 2 n→∞ n→∞ n + sin(n)
Solution 2: Divide the top and bottom by the largest power of n. s 2 an = 1 + sin(n) n Since sin(n) is bounded, clearly
sin(n) n
→ 0 as n → ∞. Hence r
lim an =
n→∞
√ 2 = 2. 1+0
Note: this problem is about sequences, not series. This means we cannot use any series tools. Doing so will almost certainly give you the wrong √ answer, as well as an incorrect justification. For instance, after showing limn→∞ an = 2, you shouldP not conclude that an fails the nth Term Test and hence diverges. The nth Term Test says ∞ n=1 an diverges, but that is a different object.
Question 7: Determine which of the following series are convergent. For full credit be sure to explain your reasoning (e.g., say what test was used). (I)
∞ X cos(nπ) n=3
(II)
πn
∞ X n=3
1 n ln(n)2
∞ X n! (III) 3 n 7n n=3
(a) only (I)
(b)* only (I) and (II)
(c) only (I) and (III)
(d) only (II)
(e) only (II) and (III)
(f) only (III)
Solution for (I): Recall that cos(nπ) = (−1)n . (This is a common trick on final exams.) So this is a geometric series with r = −1/π. Since |r| < 1, the series converges. Alternatively, you can use the alternating series test since 1/π n is positive, decreasing, and has limit 0.
Solution for (II): Use the integral test: Z
dx = x ln(x)2
Hence the improper integral Z 3
∞
Z
du 1 1 =− =− 2 u u ln(x)
1 dx = 2 x ln(x) ln(3)
converges, and so does series (II). Note: you cannot use a Direct Comparison or Limit Comparison Test with a p-series here, the result will be inconclusive. Ratio and root tests are also inconclusive (this is generally true unless an exponential or factorial term is present).
Solution for (III): This series diverges. To see this, either use the ratio test or the nth Term Test. Ratio Test:
(n + 1)! n3 7n n3 · = lim =∞>1 n→∞ (n + 1)3 7n+1 n→∞ 7(n + 1)2 n! lim
nth Term Test: Make an orders of growth argument to see that lim
n! = ∞. 7n
n→∞ n3
Question 8: Find the interval of convergence for the series below. ∞ X 3n−1 (x + 2)n √ n n n 5 n=1
(a) x = −2
(b) − 11 , − 31 3
(c)* − 11 , − 13 3
(d) − 13 , − 57 5
(e) − 13 , − 57 5
(f) (−∞, ∞)
Solution: The center is at a = −2. The radius of convergence R is given by the modified ratio test 3n 5n n3/2 1 = lim n+1 · R n→∞ 5 (n + 1)( 3/2) 3n−1 3n3/2 3 = lim = n→∞ 5(n + 1)3/2 5 It just remains to check the endpoints a − R = −11/3 and a + R = −1/3. ∞ ∞ X 3n−1 (−5/3)n 1 X (−1)n √ √ = nn n 3 5 n n n=1 n=1
This is an alternating series with positive part decreasing to 0, so it converges. Alternatively, the absolute version is a p-series with p = 3/2 so it converges absolutely. ∞ ∞ X 3n−1 (5/3)n 1X 1 √ √ = nn n 3 5 n n n=1 n=1
This is a p-series with p = 3/2 so it converges too.
Question 9: Find the Taylor polynomial P3 (x) of order 3 for ln(cos2 (x)) centered at x = π and evaluate it at x = 0. Then P3 (0) = (a) 0
(b) π
(c) ln(2) − π
(d)* − π 2
(e) 2π 2
(f) ln(2) − 2π 2
Solution: To compute the order 3 Taylor polynomial, we need the first three derivatives of f (x) = ln(cos2 (x). By the chain rule f 0 (x) = −2 cos(x) sin(x)/ cos2 (x) = −2 tan(x) f 00 (x) = −2 sec2 (x) f 000 (x) = −4 sec2 (x) tan(x) Now substituting the center x = π gives 0 2 0 (x − π)1 − (x − π)2 + (x − π)3 1! 2! 3! = −(x − π)2
P3 (x) = 0 +
Hence P3 (0) = −π 2 .
Question 10: Solve the initial value problem dy = x3 cos(x) + y, dx
x
y(π) = π
Find y( π2 ). (a)
π 2
(b)
3π 2
(c)
5π 2
(d)
π2 4
(e)* π +
π2 4
(f) 2π +
π2 4
Solution: This differential equation is linear and not separable. We begin by putting it in standard form: 1 dy − y = x2 cos(x) dx x Then the integrating factor is e
R
−1/x dx
= 1/x. So the equation becomes
1 dy 1 − 2 y = x cos(x). x dx x Integrating gives y = x
Z x cos(x)
To compute this, we must use integration by parts with u = x and dv = cos(x). Hence y = x sin(x) + cos(x) + C x so y = x2 sin(x) + x cos(x) + Cx Check: y 0 = 2x sin(x) + x2 cos(x) − x sin(x) + cos(x) + C. Therefore xy 0 = x3 cos(x) + y as desired. Now we use the initial value to find C: π = 0 − π + Cπ Hence C = 2 and y = x2 sin(x) + x cos(x) + 2π. Finally, plugging in x = y(π/2) =
π2 + 0 + 2π. 4
π 2
gives
