20b_00f_1s

20b_00f_1s - Math 20B(Bender Solutions to First Exam 18 October 2000 Ive noted if the problem or a near miss is in the text 4 x2 dx by interpreting

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Math 20B (Bender) Solutions to First Exam 18 October 2000 I’ve noted if the problem or a near miss is in the text. 1. (10 pts.) Evaluate Z 2 0 p 4 - x 2 dx by interpreting it as an area. A. (p.383, Ex.4) Squaring and rearranging y = 4 - x 2 gives x 2 + y 2 = 4, a circle of radius 2 centered at the origin. The integral is the area in the first quadrant and so equals ( π 2 2 ) / 4= π . Since the problem did not ask for an exact answer, you will receive credit for a reasonable numerical evaluation. 2. (30 pts.) Evaluate the following integrals using the tools discussed in the text. Z (1 - x ) p 2 x - x 2 dx Z 2 0 | sin πx | dx. A. (p.426, #26, #39) The substitution u =2 x - x 2 converts the first to R 1 2 u 1 / 2 du = u 3 / 2 / 3+ C and so the answer is (2 x - x 2 ) 3 / 2 / C . The second integral equals R 1 0 sin πx dx - R 2 1 sin . The substitution u = gives R sin =( - cos ) + C . Thus the answer is ( - cos π + cos 0) - ( - cos π + cos 2 π ) =4 . 3. (30 pts.) Differentiate the functions F ( x )= Z x 1 p 1+ u 4 du G ( x Z 1 x 2 ln(1 - t 3 ) dt. A. Both rely on the Fundamental Theorem of Calculus. We have F 0 ( x x
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This note was uploaded on 06/17/2009 for the course MATH 20B taught by Professor Justin during the Winter '08 term at UCSD.

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