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Section6_5_review

# Section6_5_review - -cos0 29 = 2(29-1 29 29 = 2 And f ave =...

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Section 6.5: Average Value of a Function To determine the average value of a continuous function f over the interval a , b [ ] : f ave = 1 b - a f x (29 dx a b To find the x -coordinate within the interval a , b [ ] at which the average value occurs, set f ave = fx ( 29 and solve for x . Problem. Find the average value of fx ( 29 = 6 x - x 2 on the interval 0,3 [ ] . Determine where the average occurs. Solution. f ave = 1 b - 1 f x (29 dx a b = 1 3 - 0 6 x - x 2 ( 29 dx 0 3 = 1 3 6 x 2 2 - x 3 3 0 3 = 1 3 33 ( 29 2 - 1 3 3 ( 29 3 - 0 ( 29 = 1 3 27 - 9 ( 29 = 18 3 = 6 And f ave = 6 occurs when f ave = fx ( 29 , or 6 = 6 x - x 2 x 2 - 6 x + 6 = 0 x = - b ± b 2 - 4 ac 2 a = -- 6 ( 29 ± 36 - 46 ( 29 2 = 6 ± 12 2 = 6 ± 2 3 2 = 3 ± 3 so x 4 .732 or x 1 .268 Since the interval is 0,3 [ ] , f ave occurs at 1.268. Problem. Find the average value of ft ( 29 = sin t on the interval 0, π 2 . Determine where the average occurs. Solution. f ave = 1 b - a f t ( 29 dt a b = 1 π 2 - 0 sin tdt 0 π 2 = 2 π - cos t ( 29

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Unformatted text preview: --cos0 ( 29 = 2-(29--1 ( 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 b-a f x (29 dx a b ∫ = 1 10-x sin x 2 (29 dx 10 ∫ Using a u-substitution, 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 20-cos u ( 29 100 =-1 20 cos100-cos0 ( 29 = 1-cos100 20...
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