HW8 Solutions - ASSIGNMENT 8 - SOLUTIONS Problem 1. Compute...

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Unformatted text preview: ASSIGNMENT 8 - SOLUTIONS Problem 1. Compute the following indefinite integrals: t + 2 t 2 t dt, t + 2 t 2 t + 1 dt Solution. The first integral is done by separation: t + 2 t 2 t dt = t 1 / 2 dt + 2 t 3 / 2 dt = 2 3 t 3 / 2 + 4 5 t 5 / 2 + C The second integral is done by substitution. Set u = t + 1, so dt = du . t + 2 t 2 t + 1 dt = ( u- 1) + 2( u- 1) 2 u du = 2 u 2- 3 u + 1 u du The point of the substitution is to make the denominator simpler, even at the price of making the numerator more complicated. The integral in terms of u is computed along the same lines as the first integral of this problem. 2 u 2- 3 u + 1 u du = 2 u 3 / 2- 3 u 1 / 2 + u- 1 / 2 du = 2 2 5 u 5 / 2- 3 2 3 u 3 / 2 + ( 2 u 1 / 2 ) + C So t + 2 t 2 t + 1 dt = 4 5 u 5 / 2- 2 u 3 / 2 + 2 u 1 / 2 + C = 4 5 ( t + 1) 5 / 2- 2( t + 1) 3 / 2 + 2( t + 1) 1 / 2 + C. Problem 2. Compute the following indefinite integrals: (3- 4 x 2 ) 2 dx, x (3- 4 x 2 ) 2 dx Solution. Again, the first integral is done by separation: (3- 4 x 2 ) 2 dx = 9- 24 x 2 + 16 x 4 dx = 9 x- 24 x 3 3 + 16 x 5 5 + C = 16 5 x 5- 8 x 3 + 9 x + C The second can also be done by expanding into powers of x , but substitution seems quicker: put u = 3- 4 x 2 , so du =- 8 xdx . Then x (3- 4 x 2 ) 2 dx =- 1 8...
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This note was uploaded on 07/31/2011 for the course MATH 102 taught by Professor Maryamnamazi during the Spring '10 term at University of Victoria.

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HW8 Solutions - ASSIGNMENT 8 - SOLUTIONS Problem 1. Compute...

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