Quiz 8 Solution

U12 du 1 u32 c 1 4 x2 32 c 3 3 4x1 x2

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Unformatted text preview: = −1 2 ￿ −1 2 du = −2x dx, du = x dx. u1/2 du = − 1 u3/2 + C = − 1 (4 − x2 )3/2 + C. 3 3 4x(1 + x2 )3 dx. u = 1 + x2 , du = 2x dx, 2 du = 4x dx. If x = 0, then u = 1 + 02 = 1. If x = 2, then u = 1 + 22 = 5. Then ￿2 0 23 4x(1 + x ) dx = ￿5 1 ￿5 ￿ 2u3 du = 1 u4 ￿ = 1 (54 − 1) = 312. 2 2 1 MATH 1142: Quiz 8B Solutions Nathan T. Gray Fall 2013 Exercise 1. Evaluate Solution. Let ￿ 2x(x2 + 4)5 dx. u = x 2 + 4, Then ￿ 2x(x2 + 4)5 dx = ￿ Exercise 2. Evaluate du = 2x dx. u5 du...
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