1b-2009-final_exam_sample_1_solutions

# 1b-2009-final_exam_sample_1_solutions - Math 1B — Sample...

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Unformatted text preview: Math 1B — Sample Final #1 Solutions Fall Program for Freshmen 2009 Fred Bourgoin 1. Compute the indefinite integrals. (a) Z tan 3 x sec 3 x dx Solution. Use the substitution u = sec x and du = sec x tan x dx . Z tan 3 x sec 3 x dx = Z tan 2 x sec 2 x sec x tan x dx = Z (sec 2 x- 1) sec 2 x sec x tan x dx = Z ( u 2- 1) u 2 du = Z ( u 4- u 2 ) du = 1 5 u 5- 1 3 u 3 + C = 1 5 sec 5 x- 1 3 sec 3 x + C . (b) Z dx x 3 + x Solution. Partial fractions here. Z dx x 3 + x = Z dx x ( x 2 + 1) = Z A x + Bx + C x 2 + 1 dx. A little bit of algebra shows that A = 1 and B =- 1, so Z dx x 3 + x = Z 1 x- x x 2 + 1 dx = ln | x | - 1 2 ln | x 2 + 1 | + C . 2. Evaluate each definite integral, or show that it diverges. (a) Z 1 1 2- 3 x dx Solution. The integrand is not continuous at x = 2 3 , so this is an improper integral of type 2. Z 1 1 2- 3 x dx = Z 2 / 3 1 2- 3 x dx + Z 1 2 / 3 1 2- 3 x dx = lim t → 2 3- Z t 1 2- 3 x dx + lim t → 2 3 + Z 1 t 1 2- 3 x dx. 1 Let us compute the first part: lim t → 2 3- Z t 1 2- 3 x dx = lim t → 2 3-- 1 3 ln | 2- 3 x | t = lim t → 2 3- 1 3 ln 2- 1 3 ln | 2- 3 t | = + ∞ . Therefore the integral diverges. (b) Z e 1 dx x [1 + (ln x ) 2 ] Solution. Use the substitution u = ln x and du = dx x . Don’t forget to change the limits of integration. Z e 1 dx x [1 + (ln x ) 2 ] = Z 1 du 1 + u 2 = tan- 1 i 1 = tan- 1 (1)- tan- 1 (0) = π 4 . 3. Determine whether each infinite series converges absolutely, converges condition- ally, or diverges. (a) ∞ X n =1 (- 1) n n 3 5 n Solution. The ratio test works best in this case. lim n →∞ (- 1) n +1 ( n + 1) 3 5 n +1 · 5 n (- 1) n n 3 = 1 5 lim n →∞ ( n + 1) 3 n 3 = 1 5 < 1 , so the series converges absolutely. (b) ∞ X n =2 1 n ln n Solution. Use the integral test. It’s okay because f ( x ) = 1 x ln x is positive, continuous, and decreasing on the interval (2 , ∞ ). To see that it is decreasing, compute its derivative: f ( x ) =- (1 + ln x ) ( x ln x ) 2 < 0 for x > 1 e . Now let’s compute the integral....
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## This note was uploaded on 02/19/2012 for the course MATH 1 taught by Professor Wilkening during the Spring '08 term at Berkeley.

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1b-2009-final_exam_sample_1_solutions - Math 1B — Sample...

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