math16B_review_exer2_ans-1

math16B_review_exer2_ans-1 - Math 16B Sarason REVIEW...

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Math 16B Spring 2011 Sarason REVIEW EXERCISES 2 – ANSWERS AND COMMENTS 1. (a) Z tan x cos 3 x dx = 1 3 sec 3 x + C (Make the substitution u = cos x .) (b) Z tan x cos 3 x dx = - cos 3 x 3 + C (c) Z 1 0 x ( x - 1) 4 dx = 1 30 (Make the substitution u = x - 1.) (d) Z 1 0 x 2 ( x - 1) 5 dx = - 1 168 (Make the substitution u = x - 1.) (e) Z 4 0 e - x dx = 2 - 6 e - 2 (Make the substitution x = u 2 , then integrate by parts.) (f) Z 1 x π - 5 dx = 1 4 - π (g) Z 1 x - 1 . 0001 dx = 10 , 000 (h) Z 1 x - e dx = 1 e - 1 2. The general solution of y 0 = - 3 y 2 t is y = 1 2 t 3 / 2 + C , -∞ < C < , together with the constant function y = 0. The solution satisfying y (1) = 1 is y = 1 / (2 t 3 / 2 - 1). The solution satisfying y (1) = 0 is the constant solution y = 0. 3. The general solution of 2 yy 0 = - ( y 2 - 1) 2 is y = ± q 1 + 1 t + C , -∞ < C < , plus the constant functions y = 1 and y = - 1. If y (1) = - 2 then y = - q 1 + 1 t - 2 3 . If y (1) = - 1 then y is the constant solution y = - 1. 4. The general solution of
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This note was uploaded on 04/16/2011 for the course MATH 16B taught by Professor Sarason during the Spring '06 term at Berkeley.

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math16B_review_exer2_ans-1 - Math 16B Sarason REVIEW...

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