mid1g_sol - Math 125G - Spring 2002 First Mid-Term Exam...

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Unformatted text preview: Math 125G - Spring 2002 First Mid-Term Exam Solutions 1. Is 1 2 x 2 ln x- 1 4 x 2 an antiderivative of x ln x ? Explain. The derivative of 1 2 x 2 ln x- 1 4 x 2 is x ln x (details left to the reader, must be shown for credit), so, yes, 1 2 x 2 ln x- 1 4 x 2 an antiderivative of x ln x . 2. Suppose f primeprime ( x ) = 2 + e x , f prime (0) = 3 and f (0) = 2. Find f ( x ). Since f primeprime ( x ) = 2 + e x , f prime ( x ) = integraldisplay (2 + e x ) dx = 2 x + e x + C . Using the fact that f prime (0) = 3, we have f prime (0) = 0+ e + C = 1+ C = 3, so C = 2, i.e., f prime ( x ) = 2 x + e x +2. From this, we get f ( x ) = integraldisplay (2 x + e x + 2) dx = x 2 + e x + 2 x + D , where D is an arbitrary constant. Using the fact that f (0) = 2, we have 2 = f (0) = 0 + 1 + 0 + D , so D = 1. Hence, f ( x ) = 1 + 2 x + x 2 + e x . 3. Use the midpoint rule with n = 3 to approximate the integral integraldisplay 6 ln(sin x + 3) dx....
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This note was uploaded on 05/01/2008 for the course MATH 125 taught by Professor Chen during the Spring '08 term at University of Washington.

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mid1g_sol - Math 125G - Spring 2002 First Mid-Term Exam...

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