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Unformatted text preview: cos0 ( 29 = 2(291 ( 29 ( 29 = 2 And f ave = 2 occurs when f ave = fx (29 , or 2 = sin t Using a calculator, I can obtain t ≈ .690 radians, or t ≈ 39 .54 o . Problem. Find the average value of f x (29 = x sin x 2 (29 on the interval 0,10 [ ] . Solution. f ave = 1 ba f x (29 dx a b ∫ = 1 10x sin x 2 (29 dx 10 ∫ Using a usubstitution, u = x 2 du = 2 xdx , so xdx = du 2 when x = , u = 2 = when x = 10 , u = 10 2 = 100 Therefore, f ave = 1 10 sin u ⋅ du 2 100 ∫ = 1 20cos u ( 29 100 =1 20 cos100cos0 ( 29 = 1cos100 20...
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 Fall '07
 FAMIGLIETTI
 Math, Trigraph, Continuous function, π π, 09radians

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