final-sample-ans

final-sample-ans - Math 1B Sample Answers to Sample Final...

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Math 1B. Sample Answers to Sample Final Exam 1. (35 points) Find: (a). Z 2 0 p x 2 + 4 dx Substitute x = 2tan θ ; then dx = 2sec 2 θ and x 2 + 4 = 4 tan 2 θ + 4 = 2sec θ , and we have Z 2 0 p x 2 + 4 dx = Z π/ 4 0 2 sec θ · 2 sec 2 θ = 2tan θ sec θ dθ ± ± ± π/ 4 0 +2 Z π/ 4 0 sec θ dθ = 2 2 + 2 ln | sec θ + tan θ | ± ± ± π/ 4 0 = 2 2 + 2 ln( 2 + 1) , using the provided formulas #9 and #7. (b). Z e 1 (ln x ) 2 dx First substitute t = ln x , x = e t , dx = e t dt : Z e 1 (ln x ) 2 dx = Z 1 0 t 2 e t dt . Now do integration by parts with u = t 2 dv = e t dt du = 2 tdt v = e t : Z 1 0 t 2 e t dt = t 2 e t ± ± ± 1 0 - Z 1 0 2 te t dt . Doing integration by parts again with u = 2 t dv = e t dt du = 2 dt v = e t then gives Z 1 0 2 te t dt = 2 te t ± ± ± 1 0 - Z 1 0 2 e t dt = ( 2 te t - 2 e t ) ± ± ± 1 0 , and therefore Z e 1 (ln x ) 2 dx = ( t 2 e t - 2 te t + 2 e t ) ± ± ± 1 0 = e - 2 . 1
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2 (c). Z 1 / 2 0 arctan x x dx There seems to be no way to do this using ordinary integration techniques, so let’s integrate it as a series. Using the given series for the arctangent, the integral is Z 1 / 2 0 arctan x x dx = Z 1 / 2 0 1 x X n =0 ( - 1) n x 2 n +1 2 n + 1 = Z 1 / 2 0 X n =0 ( - 1) n x 2 n 2 n + 1 = " X n =0 ( - 1) n x 2 n +1 (2 n + 1) 2 # 1 / 2 0 = X n =0 ( - 1) n (2 n + 1) 2 2 2 n +1 . This expression converges and is valid because all power series used in the above com- putation have radius of convergence 1 . 2. (12 points) If the curve y = 2 x - x 2 , 0 x 1 , is rotated about the x -axis, find the area of the resulting surface. We have y 0 = 2 - 2 x 2 2 x - x 2 = 1 - x 2 x - x 2 , so ds = p ( y 0 ) 2 + 1 dx = r (1 - x ) 2 2 x - x 2 + 1 dx = r x 2 - 2 x + 1 + 2 x - x 2 2 x - x 2 dx = dx 2 x - x 2 and therefore the area is 2 π Z 1 0 y ds = 2 π Z 1 0 p 2 x - x 2 · dx 2 x - x 2 = 2 π Z 1 0 dx = 2 π . 3. (14 points) Describe how one can compute X n =1 ( - 1) n n + 1 to within 0.01. (You do not need to actually carry out the computation, but if your answer involves, say, the n th partial sum, then you should say what n is.) This is an alternating sum, with terms decreasing to zero in absolute value, so it converges, and any given partial sum is valid to within the first omitted term. So, if we want 1 / p ( n + 1) + 1 < 0 . 01 , then n
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This note was uploaded on 08/03/2010 for the course MATH 58455 taught by Professor Daniel during the Spring '09 term at Berkeley.

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final-sample-ans - Math 1B Sample Answers to Sample Final...

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