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11Breview5sol - ucsc econ/ams 11b Review Questions 5...

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ucsc econ/ams 11 b Review Questions 5 Solutions 1. Compute the following integrals a. Substitute u = x 2 + 1, du = 2 x dx , then 3 x dx = (3 / 2) du and Z 3 x dx 3 x 2 + 1 = 3 2 Z u - 1 / 3 du = 3 2 · u 2 / 3 2 / 3 + C = 9 4 ( x 2 + 1) 2 / 3 + C . b. Substitute u = x 3 +3 x 2 - 1, du = (3 x 2 +6 x ) dx , then ( x 2 +2 x ) dx = (1 / 3) du and Z ( x 2 +2 x )( x 3 +3 x 2 - 1) 3 dx = 1 3 Z u 3 du = 1 12 u 4 + C = 1 12 ( x 3 +3 x 2 - 1) 4 + C . c. Substitute v = ln x , dv = 1 x dx , then Z dx x ln x = Z 1 ln x · 1 x dx = Z dv v = ln | v | + C = ln | ln x | + C . d. Substitution does not work here (why?), but the integrand is a polynomial: Z ( x 2 + x )( x 3 + x 2 - 1) 2 dx = Z ( x 2 + x )( x 6 + 2 x 5 + x 4 - 2 x 3 - 2 x 2 + 1) dx = Z x 8 + 2 x 7 + 3 x 6 - x 5 - 4 x 4 - 2 x 3 + x 2 + x dx = x 9 9 + x 8 4 + 3 x 7 7 - x 6 6 - 4 x 5 5 - x 4 2 + x 3 3 + x 2 2 + C . e. Substitute u = x 2 + 1, giving du = x x 2 + 1 dx , (check!!), so Z 3 x · e x 2 +1 x 2 + 1 dx = 3 Z e u du = 3 e u + C = 3 e x 2 +1 + C. Sometimes the ‘ugliest’ integrals are very easy. f. Substitute u = - 0 . 05 t , du = - . 05 dt , then dt = - 20 du and Z 1000 e - 0 . 05 t dt = - 20000 Z e u du = - 20000 e u + C = - 20000 e - 0 . 05 t + C . g. Substitute u = 3 t + 1, du = 3 dt , so dt = 1 3 du and t = 1 3 ( u - 1). Then Z t 2 + 5 3 t + 1 dt = 1 3 Z 1 9 ( u - 1) 2 + 5 u du = 1 3 Z 1 9 u - 2 9 + 46 9 1 u
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