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Unformatted text preview: 20B Quiz Solutions Fall 2003 Bender Each quiz is worth 12 points. Q1. These integrals could be done using substitution; however, substitution is not covered on this quiz; therefore, the solutions are done without it. If you did use substitution, you will still get credit. Z 8 r 2 t dt = Z 8 2 t 1 / 2 dt = 2 t 1 / 2 1 / 2 fl fl fl fl 8 = 8 . At some point you must write down an antiderivative of something like at n . If you do not get that correct, you receive no credit. If you manipulate the integrand incorrectly to start with (e.g. convert it to 2 t 1 / 2 ) but integrate what you get correctly, you get 3 points. If you get to all but the 8 at the end, you get 5 points. Due to poor design of the problem the integrand, p 2 /t , is not defined at t = 0. Therefore, anyone noting this fact and saying the integral cannot be done will receive credit for the problem. Z e x +1 dx = Z e e x dx = e e x + C = e x +1 + C. Getting e x +1 (or e e x ) is worth 4 points. Having the C (even if you dont have e x +1 ) is worth 2 points. Q2 Each part is 4 points. (a) Use the substitution u = ln t to get R u du = u 2 / 2 + C = (ln t ) 2 / 2 + C . Lose 2 points for leaving answer as u 2 / 2 + C . Lose 1 point for omitting C . If you get the wrong answer for some reason but have + C , get 1 point. (b) Use the substitution 2 x 3 = u to get R ( u + 3) u 50 du , possibly with limits. The indefinite integal is u 52 / 52 + 3 u 51 / 51 + C . There are three approaches: (i) Evaluate the indefinite integral, getting (2...
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 Fall '08
 Justin
 Calculus, Integrals

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