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

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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 0 + 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 0 ln(sin x + 3) dx. Cutting the interval [0 , 6] into three equal subintervals, we find these subintervals have length 2 and have centers 1, 3 and 5. Hence the midpoint rule approximation is 2 (ln(sin 1 + 3) + ln(sin 3 + 3) + ln(sin 5 + 3)) = 6 . 40782354841865609 ....
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