# 5.3.12-eg - Z 5-1 ± 2 x-3 x 2 ² dx = 2 Z 5-1 xdx-3 Z 5-1...

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Example Find the average value of f ( x ) = 2 x - 3 x 2 on the interval [ - 1 , 5]. Solution: Recall that the average value of a function f ( x ) on the interval [ a,b ] is given by 1 b - a Z b a f ( x ) dx, so in this case we seek the value of 1 5 - ( - 1) Z 5 - 1 ± 2 x - 3 x 2 ² dx. In order to evaluate the integral, the following following formulas may be used: Z b a xdx = b 2 2 - a 2 2 , Z b a x 2 dx = b 3 3 - a 3 3 . So using linearity properties of the deﬁnite integral, it follows that
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Unformatted text preview: Z 5-1 ± 2 x-3 x 2 ² dx = 2 Z 5-1 xdx-3 Z 5-1 x 2 dx = 2 5 2 2-(-1) 2 2 !-3 5 3 3-(-1) 3 3 ! = 2 ³ 25-1 2 ´-3 125-(-1) 3 ! = 24-126 =-102 . Thus the average value of f ( x ) = 2 x-3 x 2 over the interval [-1 , 5] is given by 1 5-(-1) Z 5-1 ± 2 x-3 x 2 ² dx =-102 6 =-17 ....
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## This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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