quiz3-sol - (2-2) + A 2 (2 + 5) ⇒-7 = 0 + 7 A 2 ⇒ A 2...

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Quiz 3 – Math 2153, Calculus II – Sept. 9, 2011 1. Evaluate the integral Z t - 9 t 2 + 3 t - 10 dt Solution We use partial fractions to solve this problem. Notice that the denominator can be factored as t 2 + 3 t - 10 = ( t + 5)( t - 2) . Therefore the integrand can be written as t - 9 t 2 + 3 t - 10 = A 1 t + 5 + A 2 t - 2 (1) Now we need to compute the values of A 1 and A 2 . Notice that A 1 t + 5 + A 2 t - 2 = A 1 ( t - 2) ( t + 5)( t - 2) + A 2 ( t + 5) ( t + 5)( t - 2) = A 1 ( t - 2) + A 2 ( t + 5) ( t + 5)( t - 2) So Equation (1) can be written as t - 9 t 2 + 3 t - 10 = A 1 ( t - 2) + A 2 ( t + 5) ( t + 5)( t - 2) Compare the numerators and denominators of the above equation, we have t - 9 = A 1 ( t - 2) + A 2 ( t + 5) (2) 1. Set t = 2 in Equation (2), we have 2 - 9 = A 1
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Unformatted text preview: (2-2) + A 2 (2 + 5) ⇒-7 = 0 + 7 A 2 ⇒ A 2 =-1 2. Set t =-5 in Equation (2), we have-5-9 = A 1 (-5-2) + A 2 (-5 + 5) ⇒-14 =-7 A 1 + 0 ⇒ A 1 = 2 Plug in the values of A 1 and A 2 back into Equation (1), we have t-9 t 2 + 3 t-10 = 2 t + 5-1 t-2 Hence the integral is Z t-9 t 2 + 3 t-10 dt = Z ± 2 t + 5-1 t-2 ² dt = 2 ln | t + 5 | -ln | t-2 | + C (Or if you like, you can further simplify the answer into ) = ln ³ ³ ³ ³ ( t + 5) 2 ( t-2) ³ ³ ³ ³ + C...
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This note was uploaded on 09/22/2011 for the course MATH 2153 taught by Professor Staff during the Fall '08 term at Oklahoma State.

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