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Unformatted text preview: Math 105  Practice Midterm 1 for Midterm 2 Solutions This practice midterm may be harder and/or longer than the real midterm. Not all question will be worth the same number of points. 1. Evaluate Z e x x dx , or show that it doesnt exist. Note: this was corrected on March 16, the earlier version missed the fact that the function is not defined at x = 0 . Lets first evaluate the indefinite integral with the substitution u = x , 2 du = 1 x dx : Z e x x dx = Z 2 e u du = 2 e u = 2 e x . The improper integral is actually improper in two ways: one limit of integration is and the other is an asymptote of the function. To be able to handle that, we should split the integral up, at some arbitrary number like x = 1, and do the two resulting improper integrals. Z e x x dx = Z 1 e x x dx + Z 1 e x x dx = lim c + 2 e x 1 c + lim d  2 e x d 1 = 2 lim c + ( e 1 e c ) + lim d ( e d e 1 ) = 2 ( ( e 1 1) + (0 e 1 ) ) = 2 . 2. Solve the initial value problem y = 1 xy , y (1) = 4 . dy dx = 1 x y Z ydy = Z 1 x dx 2 3 y 3 / 2 = 2 x + C 1 y 3 / 2 = 3 x + C 2 y = (3 x + C 2 ) 2 / 3 . y (1) = 4 4 = (3 + C 2 ) 2 / 3 3 + C 2 = 4 3 / 2 = 8 C 2 = 5 y = (3 x + 5) 2 / 3 ....
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 Fall '10
 MalabikaPramanik
 Math, Calculus

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