# PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e)...

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April 10, 3031

• The Honor Code is in effect for this examination. All work is to be your own. • No calculators.

• The exam lasts for (M.C. 4 min ea. P.C. 8 min. ea. = roughly an hour). • Be sure that your name is on every page in case pages become detached. • Be sure that you have all 10 pages of the test.

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!

1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) ... 3. (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e) ... 5. (a) (b) (c) (d) (e) 6. (a) (b) (c) (d) (e) ... 7. (a) (b) (c) (d) (e)

Please do NOT write in this box. Multiple Choice 8. 9. 10. 11. Total

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Multiple Choice

1.(7 pts.) Which series below is the MacLaurin series (Taylor series centered at 0) for x2 1 + x? x2 1 + x = x2 1 − (−x) = x 2 ∞ X n=0 (−x)n= ∞ X n=0 (−1)nxn+2, for |x| < 1. (a) ∞ X n=0 x2n+2 (b) ∞ X n=0 xn+2 n + 2 (c) ∞ X n=0 (−1)nx2n (d) ∞ X n=2 (−1)nx2n−2 n! (e) ∞ X n=0 (−1)n xn+2

2.(7 pts.) Which series below is a power series for cos(√x) ?

Solution: Since cos x = ∞ X n=0 (−1)n x 2n (2n)! , we have cos(√x) = ∞ X n=0 (−1)n( √ x)2n (2n)! = ∞ X n=0 (−1)n x n (2n)! = 1 − x 2! + x2 4! − · · · . (a) ∞ X n=0 (−1)nxn n2+ 1 (b) ∞ X n=0 (−1)nxn (2n)! (c) ∞ X n=0 (−1)nxn (2n + 1)! (d) ∞ X n=0 (−1)n√xn (2n)! (e) ∞ X n=0 (−1)nx2n− 12 (2n)!

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3.(7 pts.) Calculate

lim x→0

sin(x3) − x3

x9 .

Hint: Without MacLaurin series this may be a long problem.

Since sin x = ∞ X n=0 (−1)n x 2n+1 (2n + 1)! = x − x3 3! + x5 5! − · · · , we have sin(x3) = ∞ X n=0 (−1)n x 6n+3 (2n + 1)! = x 3x9 3! + x15 5! − · · · , and lim x→0 sin(x3) − x3 x9 = limx→0 (x3 x 9 3! + x15 5! − · · · ) − x 3 x9 = limx→0 −x 9 3! + x15 5! − · · · x9 = − 1 6. (a) 0 (b) 9 7 (c) 7 9 (d) ∞ (e) − 1 6

4.(7 pts.) What is the fourth Taylor polynomial, T4(x), for cos(2x) with center a = π?

Solution: cos(2π) = 1 cos(2x)0|x=π = −2 sin(2pi) = 0 cos(2x)(2)|x=π = −4 cos(2pi) = −4 cos(2x)(3)|x=π = 8 sin(2pi) = 0 cos(2x)(4)|x=π = 16 cos(2pi) = 16 Hence the Taylor polynomial at x = π is

1 − 2(x − π)2+ 2 3(x − π) 4 1 − 2(x − π)2+ 2 3(x − π) 4

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(a) 1 − 1 2!(x − π) 2+ 1 4!(x − π) 4 (b) 1 − 1 2!x 2+ 1 4!x 4 (c) 1 − 2(x − π)2+2 3(x − π) 4 (d) 1 − 4(x − π)2+ 16(x − π)4 (e) 1 + 4(x − π)2+ 16(x − π)4

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5.(7 pts.) Evaluate the integral Z π/2 0 sin3(x) cos5(x)dx. (a) − 1 24 (b) 0 (c) 1 4 (d) 1 24 (e) π 2

6.(7 pts.) Which of the following expressions gives the partial fraction decomposition of the function

f (x) = x

2− 2x + 6 x3(x − 3)(x2+ 4)?

Solution: Since x is a linear factor of multiplicity 3, (x − 3) is a linear factor of multiplicity 1 and (x2+ 4) is an irreducible quadratic factor of multiplicity 1, then

x2− 2x + 6 x3(x − 3)(x2+ 4) = A x3 + B x2 + C x + D x − 3 + Ex + F x2+ 4 . (a) A x3 + B x2 + C x + D x − 3 + E x2+ 4 (b) A x3 + B x − 3+ Cx + D x2+ 4 (c) A x3 + B x2 + C x + D x − 3 + Ex + F x2+ 4 (d) A x3 + B x2 + C x + D x − 3 + E x + 2 + F x − 2 (e) A x3 + B x − 3+ C x2+ 4

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7.(7 pts.) Evaluate the improper integral Z ∞

4

1

(x − 2)(x − 3)dx.

Using a partial fraction expansion R 1

(x − 2)(x − 3)dx = ln | x − 3 x − 2| + C. ThereforeR4∞ 1 (x − 2)(x − 3)dx = limt→∞ln | t − 3 t − 2| − ln | 1 2| = 0 + ln 2. (a) ln 3 (b) ln12 (c) ln 2

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Partial Credit

You must show your work on the partial credit problems to receive credit!

8.(11 pts.) Find the radius of convergence and interval of convergence of the power series ∞ X n=1 (−1)n √ n (x − 3) n

Remark: The correct answer with no justification is worth 2 points.

Set an =

(−1)n √

n (x − 3)

n. Using the Ratio Text,

lim n→∞ |an+1| |an| = lim n→∞ √ n √ n + 1|x − 3| = |x − 3|.

Hence, the radius of convergence is 1, and the series converges absolutely for |x−3| < 1, or 2 < x < 4. For the end points, when x = 2, the series is

∞ X n=1 (−1)n(−1)n √ n = ∞ X n=1 1 √ n which is divergent since it is a p-series with p = 1

2 < 1; when x = 4, the series is ∞ X n=1 (−1)n(1)n √ n which is convergent since it’s an alternating series, and bn =

1 √ n is decreasing and lim n→∞ 1 √

n = 0. (See the solution to Problem #3 for details.) Hence, the interval of convergence is 2 < x ≤ 4.

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9.(11 pts.) (a) Show that

∞ X n=0 (−1)nx2n = 1 1 + x2 provided that |x| < 1. (b) Find ∞ X n=0 (−1)n (2n + 1)(√3)2n+1.

(Hint: First use term-by-term integration on the series in part (a).)

(a) Since |x| < 1 , we have |x2| < 1. Hence 1 1 + x2 = 1 1 − (−x2) = ∞ X n=0 (−x2)n = ∞ X n=0 (−1)nx2n. (b) Integrate both the left and right hands of (a) to get

Z ∞ X n=0 (−1)nx2ndx = Z 1 1 + x2dx ⇒ ∞ X n=0 Z (−1)nx2ndx = Z 1 1 + x2dx ⇒ ∞ X n=0 (−1)n x 2n+1 2n + 1 = arctan x + C. Letting x = 0, we have C = 0. Hence, we have

∞ X n=0 (−1)n x 2n+1 2n + 1 = arctan x. Let x = √1 3. We get ∞ X (−1)n √ = arctan√1 = π.

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10.(11 pts.) Find the integral

Z

3x + 1 x3+ x2dx.

Solution: Use partial fraction decomposition 3x + 1 x3 + x2 = 3x + 1 x2(x + 1) = A x + B x2 + C x + 1 = Ax(x + 1) + B(x + 1) + Cx2 x2(x + 1) . Therefore 3x + 1 = (A + C)x2+ (A + B)x + B. It follows that A + C = 0, A + B = 3, B = 1, A = 2, B = 1, C = −2, and Z 3x + 1 x3+ x2dx = Z 2 x+ 1 x2 − 2 x + 1  dx = 2 ln |x| − 1 x − 2 ln |x + 1| + C.

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11.(11 pts.) Evaluate:

Z 1 3x

3√9 − x2 dx.

Solution: Two approaches work: trigonometric substitution with x = 3 sin θ and u substitution with u = 9 − x2. The method of trigonometric substitution is outlined here, although the latter method may be somewhat easier.

Z 1 3x

3√9 − x2 dx = Z

81 sin3θ cos2θdθ (x = 3 sin θ, dx = 3 cos θdθ) =

Z

81(1 − cos2θ) sin θ cos2θdθ (u = cos θ, du = − sin θdθ) = Z 81(u4− u2)du = 81 cos 5θ 5 − 27 cos 3θ + C cos θ = 1 3 √ 9 − x2 = (9 − x 2) 5 2 15 − (9 − x 2) 3 2 + C.

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April 10, 3031

• The Honor Code is in effect for this examination. All work is to be your own. • No calculators.

• The exam lasts for (M.C. 4 min ea. P.C. 8 min. ea. = roughly an hour). • Be sure that your name is on every page in case pages become detached. • Be sure that you have all 10 pages of the test.

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!

1. (a) (b) (c) (d) (•) 2. (a) (•) (c) (d) (e) ... 3. (a) (b) (c) (d) (•) 4. (a) (b) (•) (d) (e) ... 5. (a) (b) (c) (•) (e) 6. (a) (b) (•) (d) (e) ... 7. (a) (b) (•) (d) (e)

Please do NOT write in this box. Multiple Choice 8. 9. 10. 11. Total

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