Math 12 Final Exam 2010 Answer Key

Math 12: Final Exam Name: Instructions: There are 8 questions on this exam for a total of 100 points. You may not use any outside materials (eg. notes, calculators, cell phones, etc.). You have 3 hours to complete this exam. Remember to fully justify your answers. Problem 1 (12 Points). Find the following limits: (a) limx→3 ln x−2 x2 −3x (b) limx→0 sinh x x (c) limx→0 x cot x Proof. (a) This is of the form 0 0 so we apply l’Hopital’s rule to get lim x→3 (b) This is of the form 0 0 ln x − 2 1/(x − 2) 1 = lim = . x2 − 3x x→1 2x − 3 3 so we apply l’Hopital’s rule to get sinh x 1/2(ex − e−x ) 1/2(ex + e−x ) = lim = lim = 1. x→0 x→0 x→0 x x 1 lim (c) This is of the form 0 · ∞ so we apply l’Hopital’s rule to the fraction lim x cot x = lim x→0 x→0 x 1 = lim = 1. tan x x→0 sec2 x Problem 2 (12 Points). Evaluate the following integrals: Z x+3 √ (a) dx. 9 − x2 Z dx (b) . x3 + x2 − 2x Z x sec2 x dx. (c) Proof. 1 (a) We make the trig substitution x = 3 sin θ Z Z x+3 3 sin θ + 3 √ p dx = 3 cos θdθ 2 9−x 9 − 9 sin2 θ Z = 3 sin θ + 3dθ = −3 cos θ + 3θ + C p x = − 9 − x2 + 3 arcsin + C. 3 (b) We factor and use partial fractions to see x3 1 1 A B C = = + + 2 + x − 2x x(x + 2)(x − 1) x x+2 x−1 and so we have to solve A(x + 2)(x − 1) + Bx(x − 1) + Cx(x + 2) = 1. So we choose x = 0 to get A = − 12 , x = −2 to get B = 61 , and x = 1 to get C = 13 . So we integrate Z dx dx = 3 x + x2 − 2x Z −1/2 1/6 1/3 + + dx x x+2 x−1 1 1 1 = − ln |x| + ln |x + 2| + ln |x − 1| + C. 2 6 3 (c) Applying integration by parts with u = x and dv = sec2 xdx we get Z Z 2 x sec xdx = x tan x − tan xdx = x tan x + ln |cos x| + C. Problem 3 (8 Points). For each of the following improper integrals, determine whether it converges or diverges, and if it converges, find its value. Z ∞ dx (a) . 2 − 2x + 5 x 1 Z 9 dx (b) . 4/3 0 (x − 1) Proof. (a) This is an improper integral and we need to complete the square in the denominator to get Z ∞ Z c dx dx = lim 2 x − 2x + 5 c→∞ 1 (x − 1)2 + 4 1 2 We notice that this is of the form Z lim c→∞ 1 c 1 u2 +a2 dx = lim (x − 1)2 + 4 c→∞ µ with a = 2 and u = x − 1 so we get 1 x−1 arctan 2 2 ¶c 1 = lim c→∞ So this integral converges and has value 1 c−1 1π π arctan = = . 2 2 22 4 π 4. (b) Notice that there is a discontinuity at x = 1 so we have Z 9 0 Since dx = lim c→1 (x − 1)4/3 Z Z c 0 dx + lim (x − 1)4/3 d→1 Z d 9 dx (x − 1)4/3 dx = −3(x − 1)−1/3 (x − 1)4/3 we get Z 0 9 ³ ´ dx = lim −3(c − 1)−1/3 − 3 + 4/3 c→1 (x − 1) µ lim −3(d − 1)−1/3 + d→1 3 2 ¶ =∞ so this integral diverges. Problem 4 (8 Points). Let R be the region bounded by the curves y = x2 and y = x + 2. (a) Set up (but don’t evaluate) an integral for the volume of the solid obtained by rotating R about the x-axis. (b) Set up (but don’t evaluate) an integral for the volume of the solid obtained by rotating R about the line x = 2. Proof. The intersection points of the curve and the line are x = −1 and x = 2. (a) The cross sections are washers so we have Z 2 V =π (x + 2)2 − (x2 )2 dx. −1 (b) The cross sections are cylindrical shells so we have Z V = 2π (2 − x)(x + 2 − x2 )dx. Problem 5 (10 Points). Consider the curve given by x = sin3 t and y = cos3 t from t = 0 to t = 3 π 2. ³ (a) Find the tangent line(s) to the curve at ´ √ 3 3 1 8 , 8 . (b) Find the length of the curve. Proof. We compute x0 = 3 sin2 t cos t y 0 = −3 cos2 t sin t. (a) The t value for this point occurs where cos t = 1 2 and so t = π 3. So the tangent line has slope dy y0 cos t 1 = 0 =− = −√ . dx x sin t 3 Then the line is given by 1 1 y − = −√ 8 3 which simplifies to à √ ! 3 3 x− 8 1 1 y = −√ x + . 2 3 (b) For the arclength we have Z π/2 p π/2 q s= Z 0 = Z 9 sin4 t cos2 t + 9 cos4 t sin2 tdt 9 sin2 t cos2 t(sin2 t + cos2 t)dt 0 π/2 = 3 sin t cos tdt 0 µ = 3 sin2 t 2 ¶π/2 = 0 3 . 2 Problem 6 (6 Points). Let C1 be the curve given by the polar coordinates equation r = 2 sin θ, 0 ≤ θ ≤ π, and let C2 be the curve given by the polar coordinates equation r = 1. Find the area of the region inside C1 and outside C2 . Proof. r = 1 is a circle of radius 1 centered at (0, 0) and r = 2 sin θ is a circle of radius 1 centered at (0, 1). We find the points of intersection as 2 sin θ = 1 4 and so θ = { π4 , 3π 4 }. So we have for the area Z 3π/4 1 (4 sin2 θ − 1)dθ 2 π/4 ¶ Z 3π/4 µ 1 1 = 2 (1 − cos 2θ) − dθ 2 2 π/4 Z 3π/4 1 − cos 2θdθ = 2 π/4 µ ¶3π/4 θ sin 2θ = − 2 2 π/4 A= 3π 1 π 1 + − + 8 2 8 2 π = + 1. 4 = x3 1 Problem 7 (6 Points). Find the area of the surface obtained when the curve y = + for 1 ≤ x ≤ 2 6 2x is rotated about the y-axis. Proof. We compute y0 = 1 2 1 x − 2 2 2x and so 1 + (y 0 )2 = 1 4 1 1 x + + 4 = 4 2 4x µ 1 2 1 x + 2 2 2x ¶2 So we have ¶ µ ¶2 1 2 1 1 4 1 x + 2 dx = 2π x + ln |x| 2 2x 8 2 1 1 µ ¶ µ ¶ 1 7 = 2π 2 + ln 2 − − 0 = 2π + ln 2 . 8 8 Z S = 2π µ 2 x Problem 8 (12 Points). Determine whether each series converges absolutely, converges conditionally, or diverges. Justify your answers. (a) ∞ X cos(n + 10) . n2 + 10n n=1 (b) ∞ X n! . n n2 2 n=1 5 (c) ∞ X √ n (−1) n=0 n . n+2 Proof. (a) We know that |cos x| ≤ 1 and so ¯ ∞ ¯ ∞ ∞ X X ¯ cos(n + 10) ¯ X 1 1 ¯ ¯≤ ≤ . ¯ n2 + 10n ¯ 2 + 10n 2 n n n=1 n=1 n=1 The last sequence is a p-series with n = 2 and so is convergent, so by the comparison test P∞ ¯¯ cos(n+10) ¯¯ P∞ cos(n+10) n=1 ¯ n2 +10n ¯ converges and hence n=1 n2 +10n converges absolutely. (b) Applying the ratio test we see ¯ ¯ ¯ an+1 ¯ (n + 1)! 2n n 2 ¯ = lim lim ¯¯ · ¯ n+1 2 n→∞ 2 n→∞ an (n + 1) n! 2 (n + 1)n = lim n→∞ 2(n + 1)2 1 n2 = lim =∞ n→∞ 2 n + 1 So the series diverges. (c) Applying the alternating series test we have that the series is decreasing since √ √ n+1 n an+1 = < = an n+3 n+2 and that √ n =0 n+2 so the series is convergent. However, using the limit√ comparison test with the divergent series √ ¯ P∞ 1 P∞ ¯¯ P∞ n n n ¯ √ with the series (−1) = n=0 ¯ n=1 n n=0 n+2 we get n+2 ¯ lim an = lim n→∞ n→∞ √ n √ n · n = lim =1 n→∞ n + 2 n+2 √ ¯ P∞ ¯¯ n ¯ and so both series diverge and hence n=0 ¯(−1)n n+2 ¯ is conditionally convergent. lim n→∞ Problem 9 (8 Points). Find the interval of convergence of the power series 6 ∞ X (x + 2)n . 2n ln n n=2 Proof. We apply the ratio test: ¯ ¯ ¯ ¯ ¯ (x + 2)n+1 ¯ ln(n + 1) ¯ 2n ln n ¯¯ ¯. ¯ ¯ · lim = lim |x + 2| ¯ n→∞ ¯ 2n+1 ln(n + 1) (x + 2)n ¯ n→∞ ln n ¯ We apply L’Hopital’s Rule to get ¯ ¯ ¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ln(n + 1) ¯ ¯¯ ¯ ¯ ¯ ¯ = lim ¯ x + 2 ¯ ¯ n ¯ = ¯ x + 2 ¯ . lim |x + 2| ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ n→∞ n→∞ ln n 2 n+1 2 ¯ So we have R = 2 and convergence for −4 < x < 0. We now check the end points. For x = 0 we have the series ∞ X 1 ln n n=2 P∞ P∞ and we know ln1n > n1 for all n ≥ 2. Also n=2 n1 diverges as p-series. So we have that n=2 diverges at x = 0. For x = −4 we have ∞ X (−1)n ln n n=2 (x+2)n 2n ln n so we apply the alternating series test. Since ln x is an increasing function ln1x is a decreasing function and so an+1 < an for all n ≥ 2. We also know limn→∞ ln n = ∞ and so limn→∞ ln1n = 0. So this is a convergent alternating series. Therefore, the interval of convergence is −4 ≤ x < 0 or [−4, 0). Problem 10 (6 Points). Find the Taylor series for 1 x about 1. Proof. Successively taking derivatives we see that f (n) (x) = (−1)n n! xn+1 So we get that cn = Therefore, the series is ∞ X f (n) (x) = (−1)n . n! (−1)n (x − 1)n . n=0 Problem 11 (6 Points). (a) Find a formula for the finite sum ¸ n · X k k−1 − . (Hint: Write out a few terms.) 2k − 1 2k + 1 k=1 (b) Find ∞ · X k=1 k−1 k − 2k − 1 2k + 1 ¸ . 7 Proof. (a) We have the telescoping sum ¸ µ ¶ µ ¶ µ ¶ ¶ µ n · X k−1 k 1 1 2 2 3 n−1 n − = 0− + − + − + ··· + − 2k − 1 2k + 1 3 3 5 5 7 2n − 1 2n + 1 k=1 n =− . 2n + 1 (b) Since the sum of a series is the limit of its sequence of partial sums we have ¸ ∞ · X k−1 k n 1 − = lim − =− . n→∞ 2k − 1 2k + 1 2n + 1 2 k=1 Problem 12 (10 Points). Use power series to estimate R 1/2 0 ln(1+x) dx x to within 1/100. Proof. Recall that the MacLaurin series for ln(1 + x) is given by ln(1 + x) = ∞ X n=1 so we have (−1)n+1 xn x2 x3 =x− + − ··· n 2 3 ∞ ln(1 + x) X x x2 xn−1 = =1− + − ··· . (−1)n+1 x n 2 3 n=1 Evaluating the integral yields Z 1/2 0 ln(1 + x) = x à ∞ X xn (−1)n+1 2 n n=1 !1/2 à 0 ∞ X x2 x3 x− + − ··· 4 9 n=1 !1/2 0 ∞ X 1 = (−1)n+1 n 2 2 n n=1 = 1 1 1 1 − + − + ··· . 2 16 72 400 Since this is an alternating series, the remainder |Rn | ≤ an+1 so we need the first 3 terms which is Z 0 1/2 ln(1 + x) 1 1 1 ≈ − + . x 2 16 72 8