11Brev2sol

11Brev2sol - ucsc econ/ams 11b Review Questions 2 Solutions...

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Unformatted text preview: ucsc econ/ams 11b Review Questions 2 Solutions 1. Compute the following integrals a. Substitute u = x 2 + 1, du = 2 xdx , then 3 xdx = (3 / 2) du and Z 3 xdx 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 + xdx = 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 s = at + b , ds = adt , then dt = (1 /a ) ds and Z e at + b dt = 1 a Z e s ds = e s a + C = e at + b a + C ....
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11Brev2sol - ucsc econ/ams 11b Review Questions 2 Solutions...

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