20b_04f_fs

20b_04f_fs - Math 20B Final Exam Solutions Fall 2004 1. (a)...

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Unformatted text preview: Math 20B Final Exam Solutions Fall 2004 1. (a) Use the substitution x = u 2 so that dx = 2 u du . Then integraldisplay 9 dx 1 + x = integraldisplay 3 2 u du 1 + u = 2 integraldisplay 3 parenleftbigg 1- 1 1 + u parenrightbigg du = 2( u- ln | 1 + u | ) bracketrightBig 3 = 2(3- ln 4) . (b) This can be done using the identity cos 2 x = cos(2 x )+1 2 and integrating by parts. It can also be done with complex numbers, which is what is done here. integraldisplay cos 2 x e x dx = integraldisplay parenleftbigg e ix + e ix 2 parenrightbigg 2 e x dx = 1 4 integraldisplay ( e (1+2 i ) x + 2 e x + e (1 2 i ) x ) dx = e (1+2 i ) x 4(1 + 2 i ) + e x 2 + e (1 2 i ) x 4(1- 2 i ) + C. Since you can leave complex numbers in your answer, you can stop here. (c) You can use trig substitution, but it is much easier to set 1- x 2 = t so that- 2 x dx = dt and then we have integraldisplay x radicalbig 1- x 2 dx =- 1 2 integraldisplay t dt =- 1 2 t 3 / 2 3 / 2 + C =- (1- x 2 ) 3 / 2 3 + C....
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This note was uploaded on 06/17/2009 for the course MATH 20B taught by Professor Justin during the Fall '08 term at UCSD.

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20b_04f_fs - Math 20B Final Exam Solutions Fall 2004 1. (a)...

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