Calculus Midterm Exam 2, sample 2 - Answer Key

Math 121 Midterm Exam #2, sample 2 Answer Key 1. Compute the following integral. Z Z x+7 x+7 (a) dx = dx 3 x + 7x x(x2 + 7)   Z  Z  1 −x + 1 1 x 1 = dx = dx + 2 − + x x +7 x x2 + 7 x2 + 7   ln |x2 + 7| x 1 +C = ln |x| − + √ arctan √ 2 7 7 Partial Fractions Decomposition: x+7 A Bx + C = + 2 2 x(x + 7) x x +7 Clearing the denominator yields: x + 7 = A(x2 + 7) + (Bx + C)x x + 7 = (A + B)x2 + Cx + 7A so that A + B = 0, C = 1 and 7A = 7 Solve for A = 1, C = 1 and B = −1 Z ∞ 1 dx = lim t→∞ x ln x Z ln t (b) 2 = lim t→∞ ln 2 Z 2 t 1 dx x ln x 1 t du = lim [ln |ln |x||]ln ln 2 t→∞ u = lim (ln(ln t) − ln(ln 2)) = ∞ − ln(ln 2) = ∞ (diverges) t→∞ u = ln x x = t ⇒ u = ln t du = x1 dx x= Substitute Z ∞ (c) 8 1 2 ⇒ u = ln 1 dx = lim 2 t→∞ x − 10x + 28 Z 8 t 1 2
1 dx (x − 5)2 + 3 u=x−5 x=8⇒u=3 du = dx x=t⇒u=t−5 complete the square Substitute t−5   1 1 u = lim du = lim √ arctan √ t→∞ 3 t→∞ u2 + 3 3 3       √ 1 t−5 = lim √ arctan √ − arctan 3 t→∞ 3 3 Z t−5 3 1 1 = lim √ t→∞ 3   arctan t−5 √ 3   − arctan 3 √ 3    1 π π  1 π  π 1 3π 2π √ √ √ = = − = − = √ 2 3 6 6 6 3 3 3 6 3 Z   1 1 x using the formula dx = arctan +C 2 2 x +a a a 2. (a) Determine and state whether the following sequence converges or diverges. If it converges, compute its limit. Justify your answer. Do not just put down a number. Solution: Switch to the variable x and consider the function √ 2x4 + 5x3 + 7 . f (x) = 1 + 5x2 The limit of the sequence is equal to limit of this function, as x goes to infinity. This is √ lim x→∞ 2x4 + 5x3 + 7 1 + 5x2 √ 2x4 + 5x3 + 7 1/x2 · x→∞ 1 + 5x2 1/x2 p 2 + 5/x + 7/x4 = lim x→∞ 1/x2 + 5 √ 2+0+0 = 0+5 √ 2 = 5 = lim √ So the limit of the sequence is 2 5 . (b) Determine and state whether the following series converges or diverges. Justify your answer. ∞ √ X 2n4 + 5n3 + 7 1 + 5n2 n=1 Solution: √ We found in part (a) that the limit of the terms of this sequence is series diverges by NTDT. 2 2 5 . This is not zero, so the 3. Find the sum of the following series (which does converge). ∞ X 5n+1 32n−1 (−1)n n=1 ∞ X 52 53 54 (−1)n 5n+1 = − + 3 − 5 + ... 32n−1 3 3 3 n=1 Here we have a geometric series with a = − |r| = − 5 5 = < 1. 9 9 25 5 5 and r = − 2 = − . Note, it does converge since 3 3 9 a As a result, the sum is given by SUM= = 1−r 25 25 − − 3 3 = − 25 · 9 = − 75  = 14 5 3 14 14 1− − 9 9 ∞ X n 4. Use the Integral Test to determine and state whether the series e3n n=1 diverges. Justify all of your work. x Consider the related function f (x) = 3x with e 1. f (x) continuous for all x converges or 2. f (x) positive for x > 0 3. f (x) decreasing because f 0 (x) = e3x (1) − x 3e3x (e3x )2 Check the improper integral Z t Z ∞ 1 x xe−3x dx = lim − xe−3x dx = lim 3x t→∞ t→∞ e 3 1 1 t
t = 1 + 3 1 1 − 3x 1 < 0 when x > . e3x 3 Z t e−3x dx 1 t 1 1 = lim − xe−3x − e−3x t→∞ 3 9 1 1 ∞     1 t ∞ 1 = lim − − − − t→∞ 3 e3t 3e3   1 1 1 L’H = lim − + 3+ 3 = t→∞ 9e3t 3e 9e   1 1 − − 9e3t 9e3 3 1 4 + = 3 9e3 9e3 9e The improper integral converges, and therefore the original series Converges by the Integral Test (IT). u=x Or more simply O.S. CONV by IT. IBP: dv = e−3x dx 1 du = dx v = − e−3x 3 3 5. Determine whether each of the following series converges or diverges. Name any convergence test(s) you use, and justify all of your work. ∞ X   1 (a) n sin Diverges by nth term Divergence Test n n=1     1 ∞·0 1 since lim n sin = lim x sin n→∞ x→∞ n x      1 1 1   − 2 sin cos 1 x L0 H x x = 1 6= 0 = lim = lim = lim cos 1 1 x→∞ x→∞ x→∞ x − 2 x x  X ∞  ∞ ∞ X X 3 sin2 (3n) 3 sin2 (3n) (b) + = + n3 3n n3 3n n=1 n=1 ∞ X 1 is a Constant multiple of a Convergent p-Series with p = 3 > 1 n3 n=1 n=1 ∞ X sin2 (3n) and is therefore convergent, and is convergent by CT because the terms are bounded 3n First note that 3 =3 n3 n=1 ∞ X n=1 sin2 (3n) 1 < n and n 3 3 ∞ X n=1 1 1 is convergent geometric series with |r| = < 1. n 3 3 Finally, the original series is Convergent because it is the sum of two convergent series. 6. In each case determine whether the given series is absolutely convergent, conditionally convergent, or diverges. Name any convergence test(s) you use, and justify all of your work. (a) ∞ X n2 + 7 n7 + 2 (−1)n n=1 First examine the absolute series ∞ X n2 + 7 n7 + 2 n=1 ∞ ∞ ∞ X n2 + 7 X n2 X 1 Note that ≈ = which is a convergent p-series with p = 5 > 1. Next, n7 + 2 n7 n5 n=1 n=1 n=1 n2 +7 1+ 7 5 + 2 = lim n + 7n = lim 1 n→∞ n7 + 2 n→∞ 1+ 5 n 7 2 n Check: lim = 1 which is finite and non-zero (0 < 1 < ∞). 2 n→∞ n7 ∞ X n2 + 7 Therefore, these two series share the same behavior, and the absolute series is also n7 + 2 n=1 Convergent, by Limit Comparison Test (LCT). (Or more simply A.S CONV by LCT) Finally, we n7 4 have Absolute Convergence (A.C.) . (Not needed here but Note: this implies that the Original Series is Convergent by ACT.) (b) ∞ X (−3)n (n!)2 (2n)! n=1 Try Ratio Test: L = lim n→∞ (2n)! an+1 (−3)n+1 ((n + 1)!)2 (2n)! (−3)n+1 ((n + 1)!)2 · = lim · = lim n→∞ n→∞ an (2(n + 1))! (−3)n (n!)2 (−3)n (n!)2 (2n + 2)! = lim (−3)(n + 1)2 n→∞ (n + 1)2 1 3 n+1 3 = 3 lim = lim = n→∞ (2n + 1) · 2(n + 1) (2n + 1)(2n + 2) 2 n→∞ 2n + 1 4 (The last step is justified by, for example, l’Hôpital’s rule.) This limit L is less than 1, so the original series is Absolutely Convergent (A.C.) by the Ratio Test . (Or more simply O.S. AC by RT) (c) ∞ X n=1 1 (−1)n+1 √ n+4 First, we show the absolute series is divergent. Note that ∞ X n=1 ∞ √ X 1 1 √ which is a divergent ≈ n + 4 n=1 n 1 p-series with p = < 1. Next, 2 1 √ √ n 1 n+4 √ Check: lim = lim = lim = 1 which is finite and non-zero. Therefore, 1 4 n→∞ n→∞ n→∞ n + 4 √ 1+ √ n n   ∞ X 1 1 √ is the divergent p-Series p = < 1 , these two series share the same behavior. Since 2 n n=1 ∞ X 1 √ then is also divergent by Limit Comparison Test. As a result, we have no chance for n +4 n=1 Absolute Convergence. (Or more simply, A.S. DIV by LCT.) Secondly, we are left to examine the original alternating series with the Alternating Series Test. 1 •bn = √ >0 n+4 • lim √ n→∞ • 1 =0 n+4 1 1 1 1 < because bn+1 = √ <√ = bn bn+1 bn n +4 n+1+4 5 OR to show terms decreasing, could also show that for f (x) = √ we have f 0 (x)  =− 1 √ √ 2 x( x + 4)2 1 , x+4  < 0. Therefore, the original series converges by the Alternating Series Test. (Or more simply, O.S. CONV by AST.) Finally, we can conclude the original series is Conditionally Convergent (C.C.) 6