6.5 - , it must be the case that c = 2 . Example 2...

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6.5: Average Value of a Function (Dated: October 10, 2011) Let f be a function on an interval [ a , b ]. Dividing [ a , b ] into n subintervals of equal width Δ x = b - a n and choosing a sample point from each interval yields the average of the numbers f ( x * 1 ), f ( x * 2 ), ··· , f ( x * n ): 1 n n X i = 1 f ( x * i ) = f ( x * 1 ) + f ( x * 2 ) +···+ f ( x * n ) n . Here x * i is a sample point from the i th subterval [ x i - 1 , x i ]. Using the relation Δ x = b - a n , the average can be expressed as Δ x b - a n X i = 1 f ( x * i ) = 1 b - a n X i = 1 f ( x * i ) Δ x . Passing to the limit n →∞ yields f ave = 1 b - a Z b a f ( x ) dx , the average value of f on the interval [ a , b ]. Theorem 1 (Mean Value Theorem for Integrals) If f is con- tinuous function on [ a , b ] , then there exists a number c in [ a , b ] such that f ( c ) = f ave = 1 b - a Z b a f ( x ) dx . In other words, Z b a f ( x ) dx = f ( c )( b - a ). Example 1 (Exercise 6.5.9 in the text) Let f ( x ) = ( x - 3) 2 on [2,5] . (a) Find the average value of f on the given interval. (b) Find c such that f ave = f ( c ) . Solution. (a) By definition, f ave = 1 5 - 2 Z 5 2 ( x - 3) 2 dx = 1 3 Z 5 2 ( x 2 - 6 x + 9) dx = 1 3 x 3 3 - 6 x 2 2 + 9 x 5 2 = 1. (b) Let c be a number in [2,5] such that f ( c ) = f ave = 1 . Then ( c - 1) 2 = 1 or equivalently c - 1 1 and c = 1 ± 1 = 2,0 . Since 0 is not in [2,5]
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Unformatted text preview: , it must be the case that c = 2 . Example 2 (Exercise 6.5.13 in the text) If f is continuous and R 3 1 f ( x ) dx = 8 , show that f takes on the value 4 at least once on teh interval [1,3] . Solution. By the mean value theorem, there is a number c in [1,3] such that f ( c ) = f ave = 1 3-1 Z 3 1 f ( x ) dx = 1 2 8 = 4. Example 3 (Imitation of exercise 6.5.14 in the text) Find the numbers b such that the average value of f ( x ) = 6 x 2 + 10 x-8 on the interval [0, b ] is equal to 4 , where b > . Solution. The average value of f on the interval [0, b ] is given by f ave = 1 b-Z b (6 x 2 + 10 x-8) dx = 2 b 2 + 5 b-8. Thus f ave = 4 if and only if 2 b 2 + 5 b-8 = 4 or 2 b 2 + 5 b-12 = (2 b-3)( b + 4) = . This implies that either b = 3 2 ....
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This note was uploaded on 11/11/2011 for the course MATH 152.01 taught by Professor Geline during the Fall '09 term at Ohio State.

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