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math111_2005fall_prelim3sol

# math111_2005fall_prelim3sol - Math 111 Prelim 3 Name...

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Math 111 (Fall 2005) Prelim 3 (11/29/2005) 2 1. [16pts (4pts each)] (a) If x > 0, find the most general antiderivative of f ( x ) = 2 x 3 + 3 x 2 + 5. 2 x 4 4 + 3 x 1 1 + 5 x + C = x 4 2 3 x + 5 x + C (b) Compute integraldisplay π/ 4 0 parenleftbigg 2 sin θ + 1 cos 2 θ parenrightbigg integraldisplay π/ 4 0 parenleftbigg 2 sin θ + 1 cos 2 θ parenrightbigg = integraldisplay π/ 4 0 ( 2 sin θ + sec 2 θ ) = 2 cos θ + tan θ | π/ 4 0 = 2 cos( π/ 4) + tan( π/ 4) ( 2 cos(0) + tan(0)) = ( 2) 2 2 + 1 + 2 0 = 2 + 3
Math 111 (Fall 2005) Prelim 3 (11/29/2005) 3 (c) If x < 0, evaluate the indefinite integral integraldisplay x 3 + 1 x 4 dx integraldisplay x 3 + 1 x 4 dx = integraldisplay 1 x + 1 x 4 dx = ln | x | + x 3 3 + C = ln | x | − 1 3 x 3 + C (d) Compute integraldisplay 4 0 x e x 2 dx Let u = x 2 . Then du = 2 x dx , and so (1 / 2) du = x dx . Therefore, integraldisplay 4 0 x e x 2 dx = 1 2 integraldisplay x =4 x =0 e u du = 1 2 integraldisplay 16 0 e u du = 1 2 [ e u ] 16 0 = 1 2 ( e 16 e 0 ) = 1 2 ( e 16 1) CONTINUE TO NEXT PAGE

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Math 111 (Fall 2005) Prelim 3 (11/29/2005) 4 2. [6 pts] Assuming that f ( t ) is differentiable for all t , compute integraldisplay x 0 2 tf ( t 2 ) dt Let u = t 2 . Then du = 2 t dt . Therefore, integraldisplay x 0 2 tf ( t 2 ) dt = integraldisplay t = x t =0 f ( u ) du = integraldisplay x 2 0 f ( u ) du = f ( x 2 ) f (0) (by FTC part 2) 3. [6 pts] Use the definition of the definite integral to compute the numeric value of lim n →∞ n summationdisplay i =1 parenleftbigg 4 n
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math111_2005fall_prelim3sol - Math 111 Prelim 3 Name...

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