MIT18_014F10_pset9sols

# MIT18_014F10_pset9sols - 18.014 Problem Set 9 Solutions...

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Unformatted text preview: 18.014 Problem Set 9 Solutions Total: 24 points Problem 1: Integrate (a) dx . ( x 2 − 4 x + 4)( x 2 − 4 x + 5) (b) dx . 4 2 x − 2 x Solution (4 points) (a) We use the method of partial fractions to write 1 A B ( Cx + D ) = + + . ( x − 2) 2 ( x 2 − 4 x + 5) x − 2 ( x − 2) 2 x 2 − 4 x + 5 Making a common denominator yields the expression A ( x − 2)( x 2 − 4 x + 5) + B ( x 2 − 4 x + 5) + ( Cx + D )( x − 2) 2 = 1 . Plugging in x = 2, we get B = 1. Taking the derivative and plugging in x = 2 yields A = 0. Solving for C and D , we have ( x 2 − 4 x + 5) + ( Cx + D )( x − 2) 2 = 1. Comparing x 3 terms, we see C = 0. Comparing x 2 terms, we see D = − 1. Thus, we have dx dx dx 2 2 2 ( x − 4 x + 4)( x − 4 x + 5) = ( x − 2) 2 − x − 4 x + 5 . (Give yourself a pat on the back if you noticed from the start that the terms x 2 − 4 x +5 and x 2 − 4 x − 4 differed by 1 and got this decomposition without solving for A , B , C , and D ). Now the first integral is − ( x − 2) − 1 + C . To do the second integral, we complete the square and substitute u = x − 2 dx dx dx x 2 − 4 x + 5 = ( x − 2) 2 + 1 = u 2 + 1 = arctan u + C = arctan ( x − 2) + C. Thus, our final answer is dx ( x 2 − 4 x + 4)( x 2 − 4 x + 5) = − ( x − 2) − 1 − arctan ( x − 2) + C. 1 (b) Hopefully, after doing part (a), everyone observed 1 1 1 1 1 1 1 1 2 2 2 2 x 2 ( x − 2) = 2 x − 2 − x and x = 2 √ 2 x − √ 2 − x + √ 2 . − 2 If not, one can figure this out using partial fractions. Now, we have dx 1 1 1 1 x 4 − 2 x 2 = 4 √ 2 x − √ 2 − x + √ 2 − 2 x 2 dx = 4 √ 1 2 log | √ 2 | − log | x + √ 2 | + 2 1 x + C. x − 1 t Problem 2: Let A = e dt . Express the values of the following integrals in t +1 terms of A : a 1 1 t 1 e − t te t 2 e (a) dt (b) dt (c) dt (d) e t log(1 + t ) dt . a − 1 t − a − 1 t 2 + 1 ( t + 1) 2 Solution (4 points) For part (a), we substitute t = − u + a to get 1 u e u − a e − 1 − u − 1 du = − e − a u + 1 du = − e − a A. For part (b), we substitute u = t 2 to get 1 u 1 e 1 du = A. 2 u + 1 2 For part (c), we integrate by parts to get t 1 t e 1 e e − ( t + 1) + t + 1 dt = − 2 + 1 + A. For part (d), we integrate by parts to get 1 t e t log(1 + t ) 1 − e dt = e log(2) − A. 1 + t x Problem 3: Let F ( x ) = f ( t ) dt . Determine a formula (or formulas) for comput- ing F ( x ) for all real x if f is defined as follows: 2 ) 2 (a) f ( t ) = ( t + | t | 1 − t 2 ....
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MIT18_014F10_pset9sols - 18.014 Problem Set 9 Solutions...

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