20b_01w_1s

# 20b_01w_1s - Math 20B First Exam Solutions 4:40 PM 1(48 pts...

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Unformatted text preview: Math 20B First Exam Solutions 4:40 PM January 29, 2001 1. (48 pts.) Evaluate the following. Remember to show your work! (a) lim x → cos x- 1 e x- 1 . A . This has the form 0 / 0, so apply l’Hospital’s rule to get lim x →- sin x e x = 0 / 1 = 0. (You cannot apply l’Hospital’s rule a second time.) (b) F ( x ) given that F ( x ) = Z 2 √ x cos( t 2 ) dt. A . dF/dx = ( dF/d √ x )( d √ x/dx ) =- cos ‡ ( √ x ) 2 · (1 / 2 √ x ) =- (cos x ) / 2 √ x . (c) [5.5, #27] Z e t √ 1 + e t dt. A . Let 1 + e t = u and so e t dt = du . The integral becomes Z u 1 / 2 du = 2 u 3 / 2 / 3 + C = 2(1 + e t ) 3 / 2 / 3 + C. You could also have used the substitution u = e t followed by 1 + u = v . (d) Z 2 | x- 1 | dx. A . This equals Z 1 (1- x ) dx + Z 2 1 ( x- 1) dx , which equals 1 after a little bit of work. Alternatively, you could have looked at the graph and seen that the integral was the area of two right triangles with height and base 1....
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20b_01w_1s - Math 20B First Exam Solutions 4:40 PM 1(48 pts...

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