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HW2 - y tan θ-0 = tan(tan θ ⇒ d dθ ± R tan θ sec 2...

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Math 191 Problem Set 2 Solutions 5.3: 2. R 0 - 1 2 x 3 dx 5.3: 6. R 1 0 4 - x 2 dx 5.3: 14. (a) R 3 1 h ( r ) dr = R 3 - 1 h ( r ) dr - R - 1 1 h ( r ) dr = 6 - 0 = 6 (b) - R 1 3 h ( u ) du = - ( - R 3 1 h ( u ) du ) = R 3 1 h ( u ) du = 6 5.3: 22. y = 1 + 1 - x 2 y - 1 = 1 - x 2 ( y - 1) 2 = 1 - x 2 x 2 + ( y - 1) 2 = 1, a circle with center (0, 1) and radius of 1 y = 1 + 1 - x 2 is the upper semicircle. The area of this semicircle is A = 1 / 2 · πr 2 = 1 / 2 · π (1) 2 = π/ 2. The area of the rectangular base is A = lw = 2 · 1 = 2. Then the total area is 2 + π/ 2 R 1 - 1 ( 1 + 1 - x 2 ) dx = 2 + π/ 2 square units. -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x f(x) = 1+sqrt(1-x 2 ) 5.3: 60. av ( f ) = 1 1 - ( - 2) Z 1 - 2 ( t 2 - t ) dt (1) = 1 3 Z 1 - 2 t 2 dt - 1 3 Z 1 - 2 tdt (2) = 1 3 Z 1 0 t 2 dt - 1 3 Z 0 - 2 t 2 dt - 1 3 ( 1 2 / 2 - ( - 2) 2 / 2 ) (3) = 1 / 3(1 3 / 3) - 1 / 3(( - 2) 3 / 3) + 1 / 2 = 3 / 2 (4) 1
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5.4: 14. R π/ 3 0 4 sec u tan udu = [4 sec u ] π/ 3 0 = 4 sec( π/ 3) - 4 sec 0 = 4(2) - 4(1) = 4 5.4: 24. R 4 9 1 - u u du = R 4 9 ( u - 1 / 2 - 1) du = [2 u - u ] 4 9 = (2 4 - 4) - (2 9 - 9) = 3 5.4: 28. (a) R sin x 1 3 t 2 dt = [ t 3 ] sin x 1 = sin 3 x - 1 d dx ± R sin x 1 3 t 2 dt ² = d dx (sin 3 x - 1) = 3 sin 2 x cos x (b) d dx ± R sin x 1 3 t 2 dt ² = (3 sin 2 x ) d dx (sin x ) = 3 sin 2 x cos x 5.4: 30. (a) R tan θ 0 sec 2 ydy = [tan
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Unformatted text preview: y ] tan θ-0 = tan(tan θ ) ⇒ d dθ ± R tan θ sec 2 ydy ² = d dθ (tan(tan θ )) = (sec 2 (tan θ )) sec 2 θ (b) d dθ ± R tan θ sec 2 ydy ² = (sec 2 (tan θ )) d dθ (tan θ ) = (sec 2 (tan θ )) sec 2 θ 5.4: 36. y = R tan x dt 1+ t 2 ⇒ dy dx = ± 1 1+tan 2 x ² ( d dx (tan x ) ) = (1 / sec 2 x )(sec 2 x ) = 1 5.4: 43. The area of the rectangle bounded by the lines y = 2, y = 0, x = π , and x = 0 is 2 π . The area under the curve y = 1 + cos x on [0 ,π ] is R π (1 + cos x ) dx = [ x + sin x ] π = ( π + sin π )-(0 + sin 0) = π . Therefore the area of the shaded region is 2 π-π = π . 2...
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HW2 - y tan θ-0 = tan(tan θ ⇒ d dθ ± R tan θ sec 2...

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