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7thsol - Calculus 119 7th Week1 1 LowerEslimate = L5 = 5...

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Calculus 119 7th Week 1 1. LowerEslimate = L 5 = 5 i =1 Δ xf ( x i - 1 ) = Δ x ( f ( x 0 ) + f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 4 )) = 5( f (0) + f (5) + f (10) + f (15) + f (20)) = 5( - 42 - 37 - 25 - 6 + 15) = - 475 UpperEstimate = R 5 = 5 i =1 Δ xf ( x i ) = Δ x ( f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 4 ) + f ( x 5 )) = 5( f (5) + f (10) + f (15) + f (20) f (25)) = 5( - 37 - 25 - 6 + 15 + 36) = - 85 2. g ( x ) = R x 2 tan x 1 2+ t 4 dt = R 0 tan x 1 2+ t 4 dt + R x 2 0 1 2+ t 4 dt = - R tan x 0 1 2+ t 4 dt + R x 2 0 1 2+ t 4 Let u = u ( x ) = tan x and v = v ( x ) = x 2 g ( x ) = - R u 0 1 2+ t 4 dt + R v 0 1 2+ t 4 g 0 ( x ) = - 1 2+ u 4 u 0 ( x ) + 1 2+ v 4 v 0 ( x ) = - 1 2+(tan x ) 4 sec 2 x + 1 2+( x 2 ) 4 2 x = - sec 2 x 2+tan 4 x + 2 x 2+( x ) 8 3. I = lim n →∞ 1 n ( s 1 n + s 2 n + ... + r n n ) =? 1 Please email all corrections and suggestions to these solutions to [email protected] All solutions are available on the web at the url http://www.metu.edu.tr/ htor. 1

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Consider Δ x = 1 n , x i = 0 + i Δ x = i n and f ( x ) = x I = Z 1 0 xdx = x 3 2 3 2 1 0 = 2 3 (1 3 2 - 0) = 2 3 4. First Way Let u = 2 θ du = 2 Z sec 2 θ tan 2 θdθ = Z sec u tan u du 2 = 1 2 Z sec u tan udθ = 1 2 sec u + C = 1 2 sec 2 θ + C Second Way I = Z sec 2 θ tan 2 θdθ = Z 1 cos 2 θ sin 2 θ cos 2 θ Let cos 2 θ = u ⇒ - 2 sin 2 θdθ = du I = Z - du 2 u 2 = - 1 2 u - 1 - 1 + C = 1 2 u + C = 1 2 cos 2 θ + C = 1 2 sec 2 θ + C 5. Let u = x 3 + 1 du = 3 x 2 dx and x 3 = u - 1 Z 2 0 3 x 3 + 1 x 5 dx = Z 2 0 3 x 3 + 1 x 3 x 2 dx = Z 9 1 3 u ( u - 1) du 3 = 1 3 Z 9 1 u 4 3 - u 1 3 du = 1 3 ( u 7 3 7 3 - u 4 3 4 3 ) 9 1 = u 7 3 7 - u 4 3 4 9 1 = 9 7 3 7 - 9 4 3 4 - ( 1 7 3 7 - 1 4 3 4 ) 6. First Way
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7thsol - Calculus 119 7th Week1 1 LowerEslimate = L5 = 5...

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