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5.6.19-eg-1 - Example Evaluate the denite integral 6x 3 0 1...

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Example Evaluate the definite integral Z 0 - 1 6 x 3 5 x 4 + 12 dx. Solution: Let u = 5 x 4 + 12 = du = 20 x 3 dx so that x 3 dx = 1 2 du. The lower and upper limits of integration for x variable are 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 · 0 4 + 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
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