quiz1-sol

# quiz1-sol - Name MA 366 Spring 2010 Quiz 1(1(10 points...

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Unformatted text preview: Name: MA 366, Spring 2010, Quiz 1 (1) (10 points) Evaluate the following indefinite integral: x2 sin x dx. Solution: x2 sin x dx = x2 (- cos x) - u dv u v (- cos x) 2x dx . v du = -x2 cos x + 2 x cos x dx u dv = -x2 cos x + 2x sin x - 2 sin x dx = -x2 cos x + 2x sin x + 2 cos x + C. (2) (10 points) Evaluate the following definite integral: 1 0 s+2 ds s2 + 1 Solution: 1 0 s+2 ds = s2 + 1 = 1 0 2 1 1 2s 1 +2 2 ds 2+1 2s s +1 11 du 2u 1 (let u = s2 + 1, du = 2s ds) + 0 2 s2 1 ds +1 1 2 1 = ln u + 2 arctan s 2 0 1 1 = ln 2 - 0 + 2 - 0 2 4 1 = (ln 2 + ). 2 1 (3) (10 points) Evaluate the following indefinite integrals. (i) s ds s2 - s - 6 Solution: We see that s2 - s - 6 = (s - 3)(s + 2) and set s A B = + -s-6 s-3 s+2 s2 so s = A(s + 2) + B(s - 3). We'll choose values of s to make the coefficients of A and B zero: Let s = -2, -2 = A(-2 + 2) + B(-2 - 3) = -5B, so B = let s = 3, 3 = A(3 + 2) + B(3 - 3) = 5A, so A = So s2 s ds = -s-6 3 1 2 1 3 2 + ds = ln |s - 3| + ln |s + 2| + C. 5s-3 5s+2 5 5 2 ; 5 3 . 5 (ii) s2 1 ds + 2s + 2 Solution: s2 + 2s + 2 doesn't have real factors, so we complete the square: s2 + 2s + 2 = (s + 1)2 + 1 and 1 ds = + 2s + 2 1 ds (s + 1)2 + 1 1 du, where u = s + 1 and du = ds, = u2 + 1 = arctan u + C = arctan(s + 1) + C. s2 2 ...
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## This note was uploaded on 01/18/2012 for the course MATH 366 taught by Professor Edraygoins during the Fall '09 term at Purdue.

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