2005a_x1a_sols

2005a_x1a_sols - Spring 2005 Math 152 Exam 1A: Solutions c...

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Spring 2005 Math 152 Exam 1A: Solutions Mon, 21/Feb c ± 2005, Art Belmonte For specificity, lengths are in centimeters unless stated otherwise. Graphs appear at the bottom of the right column to conserve space. 1. (b) We have f ave = 1 b - a R b a f ( x ) dx = 1 π - 0 R π 0 x sin ( x 2 ) = ± - 1 2 π cos ( x 2 ) ² ³ ³ π 0 = ± 1 2 · 1 π ² - ± - 1 2 · 1 π ² = 1 π . 2. (d) Via Hooke’s Law we have F ( x ) = kx or 3 = k ± 1 6 ² , whence k = 18. Thus W = R 2 0 18 xdx = 9 x 2 ³ ³ 2 0 = 36 ft-lb. 3. (a) The area is A = R 2 0 ³ ³ x 2 - 1 ³ ³ - 0 , computed thus: A = Z 1 0 1 - x 2 + Z 2 1 x 2 - 1 = ± x - 1 3 x 3 ² ³ ³ 1 0 + ± 1 3 x 3 - x ² ³ ³ 2 1 = ± 2 3 ² - ( 0 ) + ± 8 3 - 2 ² - ± 1 3 - 1 ² = 4 3 + 8 3 - 2 = 2cm 2 . 4. (e) Use half-angle identities. R π/ 2 0 sin 2 x cos 2 = R 2 0 1 2 ( 1 - cos 2 x ) · 1 2 ( 1 + cos 2 x ) = 1 4 R 2 0 1 - cos 2 2 = 1 4 R 2 0 1 - 1 2 ( 1 + cos 4 x ) = 1 4 · 1 2 R 2 0 1 - cos 4 = 1 8 ± x - 1 4 sin 4 x ² ³ ³ 2 0 = π 16 - 0 = π 16 . 5. (c) We have dV = Adx = π r 2 = π y 2 = π ( e x ) 2 = π e 2 x . Therefore V = R 3 0 π e 2 x = 1 2 π e 2 x ³ ³ 3 0 = π 2 ( e 6 - 1 ) 621 . 13 cm 3 . 6. (a) First compute an antiderivative, then apply the FTC. Let u = ln xd v = du = 1 x v = 1 2 x 2 . Then R x ln = 1 2 x 2 ln x - R 1 2 = 1 2 x 2 ln x - 1 4 x 2 .
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2005a_x1a_sols - Spring 2005 Math 152 Exam 1A: Solutions c...

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