20b_00f_fs

20b_00f_fs - Math 20B (Bender) Final Exam Solutions Fall...

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Math 20B (Bender) Final Exam Solutions Fall 2000 1. (a) Since this is of the form 0 / 0, we apply l’Hˆ opital’s rule: lim x 0 e x - 1 sin x = lim x 0 e x cos x = 1 1 =1 . (b) You could convert to polar form: r = 2 and θ = π/ 4, giving r 30 =2 15 and 30 θ =15 π/ 2=8 π - π/ 2. Thus (1 + i ) 30 = - 2 15 i . Alternatively, you can take powers: (1 + i ) 30 = ((1 + i ) 2 ) 15 =(2 i ) 15 =2 15 i 15 =2 15 i 3 = - 2 15 i. 2. (a) The denominator factors as x ( x 2 - 1) = x ( x - 1)( x + 1). Hence 2 x 3 - x = A x + B x - 1 + C x +1 = A ( x 2 - 1) + B ( x 2 + x )+ C ( x 2 - x ) x ( x - 1)( x +1) . Thus ( A + B + C ) x 2 +( B - C ) x - A = 2. From constant and linear terms, A = - 2 and B = C . Thus the quadratic gives - 2+2 B = 0 and so B = 1. Thus we have Z 2 x 3 - x dx = Z ± - 2 x + 1 x - 1 + 1 x +1 dx =ln x 2 - 1 x 2 + C =ln | 1 - x - 2 | + C. (b) First method: Integrate by parts with u =ln x and dv = dx . This gives Z e 1 ln xdx = x ln x e 1 - Z e 1 x dx x = e - ( x e 1 = e - ( e - 1) = 1 . Second method:
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20b_00f_fs - Math 20B (Bender) Final Exam Solutions Fall...

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