prelim2_fa08_solutions - Math 1910: Prelim II Solutions...

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Unformatted text preview: Math 1910: Prelim II Solutions (Oct. 30, 2008) Problem 1. Evaluate the following indefinite integrals: (a) ( 8pts) Z log 10 ( x 2 ) x dx = Solution Z log 10 x 2 x dx = Z ln x 2 / ln 10 x dx = Z 2 ln x/ ln 10 x dx = 1 ln 10 Z 2 u du, (used substitution u = ln x ) = u 2 ln 10 + C = (ln x ) 2 ln 10 + C (b) ( 8pts) I ( x ) = Z 2- x 1- x 2 dx = Solution Note that the integrated function is defined only when- 1 < x < 1. Rewrite this as a sum of two integrals: I ( x ) = Z 2 1- x 2 dx- Z x 1- x 2 dx Use a substitution x = sin v on the 1st of them (with- 2 < v < 2 ) and u = 1- x 2 on the 2nd: I ( x ) = Z 2 cos( v ) cos( v ) dv- Z- . 5 u du = 2 v + u + C = 2 sin- 1 x + 1- x 2 + C. (c) ( 9pts) Z 8 e x- e 2 x + 2 e x + 3 dx = Hint: start with a substitution u = e x . 1 Solution Use the above substitution and partial fractions: Z 8 e x- e 2 x + 2 e x + 3 dx = Z 8- u 2 + 2 u + 3 du = Z- 8 ( u + 1)( u- 3) du = Z 2 u + 1- 2 u- 3 du = 2 ln | u + 1 | - 2 ln | u- 3 | + C = ln " u + 1 u- 3 2 # + C = ln " e x + 1 e x- 3 2 # + C. Problem 2. ( 15pts) Evaluate Z sinh x sin x dx = Solution Use integration by parts I = Z sinh x sin x dx = sinh x Z sin x dx + Z cosh x cos x dx =- sinh x cos x + cosh x Z cos x dx- Z sinh x sin x dx =- sinh x cos x + cosh x Z cos x dx- I + C =- sinh x cos x + cosh x sin x- I + C. Hence, 2 I =- sinh x cos x + cosh x sin x + C or I = 1 2 (- sinh x cos x + cosh x sin x ) + C. Problem 3. (a) ( 7pts) Show that the function f ( x ) = x e ( x 2 ) is one to one and hence has an inverse f- 1 ( x )....
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prelim2_fa08_solutions - Math 1910: Prelim II Solutions...

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