5.6.19-eg-1 - Example Evaluate the definite integral 6x 3...

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Unformatted text preview: Example Evaluate the definite integral 6x 3 0 √ −1 5x4 + 12 dx. Solution: Let u = 5x4 + 12 =⇒ du = 20x3 dx so that 1 du. 2 The lower and upper limits of integration for x variable are x3 dx = lower limit for x = −1, upper limit for x = 0. To convert this to a corresponding range of integration for the u-variable, we note that u(−1) = 5(−1)4 + 12 = 17, u(0) = 5 · 04 + 12 = 12, so that the lower and upper limits for the u variable are lower limit for u = 17, upper limit for u = 12. Although the limits of integration for x are in the usual order, i.e., −1 < 0, these need not be true for the limits of integration for u; in fact, in this case the lower limit of integration for u is larger than its upper limit. Note: One should not switch the limits of integration for the u variable without also multiplying the resulting integral by (−1) ! Making the above substitutions, 0 6x3 √ dx = 6 3x4 + 12 0 1 √ · x3 dx −1 −1 3x4 + 12 12 1 1 √· =6 du u 20 17 3 12 −1/2 = u du 10 −1 12 3 u1/2 = 10 1/2 17 √ 3√ = 12 − 17 5 √ 3√ 2 3 − 17 , = 5 which is the final answer. There is no need to return to the original variable of integration (in this case, x) for a definite integral problem. ...
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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