MATH 104 SAMPLE FINAL SOLUTIONS (1) Evaluate the integral Z e
Transcription
MATH 104 SAMPLE FINAL SOLUTIONS (1) Evaluate the integral Z e
MATH 104 SAMPLE FINAL SOLUTIONS CLAY SHONKWILER (1) Evaluate the integral Z e−x/2 cos xdx. Answer: We integrate by parts. Let u = e−x/2 and dv = cos xdx. Then du = − 12 e−x/2 dx and v = sin x. Then the above integral is equal to Z 1 e−x/2 sin x + e−x/2 sin xdx. 2 To compute this integral, again let u = e−x/2 and let dv = sin xdx. Then du = − 12 e−x/2 and v = − cos x. Hence Z Z 1 −x/2 −x/2 e cos xdx = e sin x + e−x/2 sin xdx 2 Z 1 1 −x/2 −x/2 −x/2 −e cos x − e cos xdx =e sin x + 2 2 Z 1 −x/2 1 −x/2 =e sin x − e cos x − e−x/2 cos xdx. 2 4 Therefore, Z 5 1 e−x/2 cos xdx = e−x/2 sin x − e−x/2 cos x, 4 2 so Z 4 2 e−x/2 cos xdx = e−x/2 sin x − e−x/2 cos x. 5 5 (2) Consider the series ∞ X n=1 12n . 6n2 + ln n Does this series converge or diverge? Answer: Since ln n grows significantly slower than both n and n2 , we suspect that it won’t much affect whether the series converges or P diverges. P 2Therefore, let’s do a limit comparison with the series 12n = n , which we know diverges: 6n2 12n 6n2 +ln n 2 n→∞ n lim 12n2 . n→∞ 12n2 + 2 ln n = lim 1 2 CLAY SHONKWILER By two applications of L’Hˆopital’s Rule, this limit is equal to 24n 24 lim = lim = 1. n→∞ 24n + 2 n→∞ 24 − 22 n n P2 P 12n also diverges. Therefore, since n diverges, the series 6n2 +ln n (3) Consider the sequence {an } with n(n+1) (−1) 2 5n an = . π 2n Does this sequence converge or diverge? If it converges, what does it converge to? Answer: Note that n (−1) n(n+1) 2 5n 5n 5 = . = π 2n π 2n π2 n Since π52 < 1, the sequence { π52 } converges to zero. Therefore, by the sandwich theorem, the √ given sequence converges to zero. (4) Consider the curve y = x. Find the surface area of the surface obtained by rotating the portion of the curve between x = 3/4 and x = 15/4 about the x-axis. Answer: Recall the surface area formula s 2 Z 15/4 dy 2πy 1 + dx. SA = dx 3/4 2 dy dy 1 Now, dx = 2√1 x , so dx . Hence, = 4x r Z 15/4 Z 15/4 r √ 1 1 SA = 2π x 1+ dx = 2π x + dx. 4x 4 3/4 3/4 Letting u = x + 14 , du = dx and so the above integral becomes Z 4 √ 2 3/2 4 2 2 28π = 2π (8) − (1) = . 2π udu = 2π u 3 3 3 3 1 1 (5) Suppose you invest $1000 in an account earning 8% interest compounded continuously. After how many years will you quadruple your money (use the approximation ln 4 ≈ 1.6). Answer Remember that the amount of money in your account is modeled by A(t) = P ert . Since the initial amount in the account is $1000, P = 1000. Also, r = 0.08. You will quadruple your money when 4000 = A(t) = 1000e0.08t , so 4 = e0.08t . MATH 104 SAMPLE FINAL SOLUTIONS 3 Solving for t, ln 4 = 0.08t, so ln 4 1.6 160 ≈ = = 20. 0.08 0.08 8 Therefore, you will quadruple your money in approximately 20 years. (6) Consider the series ∞ X 4n . e2n+1 t= n=1 Does this series converge or diverge? If it converges, to what sum does it converge? Answer: Note that n−1 ∞ ∞ ∞ ∞ X X X 4n 1 4n 1 4 n X 4 4 = = = . 2n+1 2n 2 3 e ee e e e e2 n=1 n=1 n=1 n=1 This is a geometric series with a = series converges to 4 e3 1− 4 e2 = 4 e3 e2 −4 e2 = 4 e3 and r = 4 e2 < 1. Hence, the 4 . e3 − 4e (7) Does the series ∞ X (2n)! n=1 2n n! converge or diverge? Answer: Use the Ratio Test: (2(n+1))! 2n+1 (n+1)! lim (2n)! n→∞ 2n n! (2n + 2)(2n + 1) 4n2 + 5n + 2 (2n + 2)! 2n n! · = lim = lim . n→∞ n→∞ n→∞ 2n+1 (n + 1)! (2n)! 2(n + 1) 2n + 1 = lim By L’Hˆopital’s Rule, this limit is equal to 8n + 5 lim = ∞, n→∞ 2 so the series diverges. (8) Find the interval of convergence of the power series ∞ X (2x)n n=0 n2 . For what values of x does the series converge absolutely? For what values does it converge conditionally? Answer: Using the Ratio Test, (2x)n+1 (n+1)2 (2x)n+1 2xn2 n2 . lim · = lim n = lim n→∞ (2x) n→∞ (n + 1)2 (2x)n n→∞ (n + 1)2 2 n 4 CLAY SHONKWILER By two applications of L’Hˆopital’s Rule, this limit is equal to 4xn = lim 4x = |2x|. lim n→∞ 2(n + 1) n→∞ 2 Now, |2x| < 1 when |x| < 12 , so the radius of convergence of the power series is 12 . Now, we need to check the endpoints, x = ± 12 . When x = − 21 , the series becomes ∞ X (−1)n n=1 n2 , which converges, by the alternating series test. In fact, ∞ ∞ X (−1)n X 1 = , n2 n2 n=1 n=1 which is a convergent p-series. On the other hand, when x = 21 , the series becomes ∞ X 1n , n2 n=1 which is a convergent p-series. Therefore, the interval of convergence of the power series is − 12 , 12 . The series converges absolutely for − 21 ≤ x ≤ 21 . 1 (9) Consider the region contained by y = 1+x 2 , the x- and y-axes and x = 1. What is the volume of the solid obtained by rotating this region about the y-axis? Answer: Using the shell method, the volume is equal to Z 1 2πrhdx. 0 1 , 1+x2 Now, r = x and h = Z 1 V = 2πx 0 so 1 dx = π 1 + x2 Z 0 1 2xdx . 1 + x2 x2 , Letting u = 1 + du = 2xdx, so we can re-write this integral as Z 2 du = π [ln |u|]21 = π [ln 2 − ln 1] = π ln 2. π 1 u (10) Evaluate the integral Z 2dx √ . (2x − 2) 4x2 − 8x + 1 Answer: We complete the square: 4x2 − 8x + 1 = (4x2 − 8x + 4) + 1 − 4 = (2x − 2)2 − 3. MATH 104 SAMPLE FINAL SOLUTIONS 5 Hence, letting u = 2x − 2, du = 2dx and so we can re-write the integral as Z du 1 1 −1 u −1 2x − 2 √ √ √ √ √ = sec +C = sec + C. 3 3 3 3 u u2 − 3 (11) Evaluate the integral Z x2 2x + 3 dx. − 7x + 12 Answer: Note that x2 − 7x + 12 = (x − 3)(x − 4). Thus, we set 2x+3 A B up the partial fraction (x−3)(x−4) = x−3 + x−4 and solve for A and B: 2x + 3 = A(x − 4) + B(x − 3). Letting x = 4, 2(4) + 3 = A(0) + B(1) = B, so B = 11. Now, letting x = 3, 2(3) + 3) = A(−1) + B(0) = −A, Z so A = −9. Therefore, Z −9 11 2x + 3 dx = + dx = −9 ln |x−3|+11 ln |x−4|+C. x2 − 12x + 7 x−3 x−4 (12) Solve the initial value problem dy 2 + y = sin x + cos x, x > 0, y(π) = . dx π Answer: Re-write in standard form: 1 sin x + cos x dy + y= , dx x x x so P (x) = 1 x x and Q(x) = sin x+cos . Hence, x Z Z 1 dx = ln |x| = ln x, P (x)dx = x since x > 0. Thus, R v(x) = e P (x)dx = eln x = x. Therefore, Z Z Z 1 1 sin x + cos x 1 y= v(x)Q(x)dx = x dx = [sin x + cos x] dx v(x) x x x 1 = [− cos x + sin x + C] x sin x − cos x + C = . x 6 CLAY SHONKWILER Using the initial value, 2 sin π − cos π + C 1+C = y(π) = = , π π π so C = 1. Therefore, y= sin x − cos x + 1 . x (13) Evaluate the integral Z (x2 x dx. + 3)3/2 √ √ Answer: Let x = 3 tan θ. Then dx = 3 sec2 θdθ, so we can re-write the integral as √ Z Z √ 3 tan θ sec2 θdθ 3 tan θ 2 3 sec θdθ = (3 tan2 θ + 3)3/2 (3 sec2 θ)3/2 Z 3 tan θ sec2 θdθ √ = 3 3 sec3 θ Z 1 tan θdθ =√ sec θ 3 Z 1 sin θdθ =√ 3 1 = − √ cos θ + C 3 Now, θ = tan−1 to √x , 3 so cos θ = √ √ 3 . 3+x2 Hence, the integral is equal √ 3 1 1 −√ √ = −√ . 2 3 3+x 3 + x2 (14) Consider the sequence {an } where √ 3 n . an = ln(n + 1) Does the sequence converge or diverge? If it converges, what does it converge to? Answer: Using L’Hˆopital’s Rule, √ 1 3 n n+1 2/3 lim = lim 3n1 = lim . n→∞ ln(n + 1) n→∞ n→∞ 3n2/3 n+1 Using L’Hˆopital again, this is equal to √ 3 1 n = ∞. lim = lim −1/3 n→∞ 2n n→∞ 2 Therefore, the sequence diverges. MATH 104 SAMPLE FINAL SOLUTIONS (15) Does the series ∞ X n=1 1 n(ln n)2 converge or diverge? Answer: Using the Integral Test, Z ∞ Z b 1 dx dx = lim . 2 b→∞ 1 x(ln x)2 x(ln x) 1 Now, letting u = ln x, du = x1 dx, so Z Z dx du 1 1 = =− +C =− + C. x(ln x)2 u2 u ln x Hence, Z b 1 1 b 1 dx = lim − = lim − + =∞ lim b→∞ b→∞ 1 x(ln x)2 ln x 1 b→∞ ln b ln 1 P 1 since ln 1 = 0. Therefore, by the Integral Test, diverges. n(ln n)2 DRL 3E3A, University of Pennsylvania E-mail address: [email protected] 7