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20b_04f_fs

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

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Math 20B Final Exam Solutions Fall 2004 1. (a) Use the substitution x = u 2 so that dx = 2 u du . Then integraldisplay 9 0 dx 1 + x = integraldisplay 3 0 2 u du 1 + u = 2 integraldisplay 3 0 parenleftbigg 1 - 1 1 + u parenrightbigg du = 2( u - ln | 1 + u | ) bracketrightBig 3 0 = 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. (d) This is an improper integral because we are dividing by zero when x = 0. Thus we need to write integraltext 1 1 = integraltext 0

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

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