MATH 12002 - CALCULUS I § 4.4: Average Value of a Function Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7
Average Value of a Function Let y = f ( x ) be a continuous function on the interval [ a , b ]. We would like to compute the average y -value of this function. Since there are infinitely many y -values (unless f is a constant function) we cannot simply add them up and divide by the number we have. We will take an approach similar to what we did to compute areas: 1 Compute a finite approximation; that is, find the average of the y -values at finitely many x -values. 2 Take the limit as the number of x -values approaches infinity. Since this is very similar to the procedure for computing definite integrals, it should not be surprising that computing average value will involve a definite integral. D.L. White (Kent State University) 2 / 7
Average Value of a Function Let y = f ( x ) be a continuous function on the interval [ a , b ]. For each positive integer n , partition the interval [ a , b ] into n equal subintervals, each of length ∆ x = b − a n , with endpoints a = x 0 < x 1 < x 2 < · · · < x n − 1 < x n = b . Compute the average of the y -values at the right endpoints of the subintervals: A n = f ( x 1 ) + f ( x 2 ) + · · · + f ( x n − 1 ) + f ( x n ) . n We define the average value f ave of f on [ a , b ] by f ave = lim n →∞ A n . D.L. White (Kent State University) 3 / 7
Average Value of a Function To compute the limit of this average value as n → ∞ , observe that f ( x 1 ) + f ( x 2 ) + · · · + f ( x n − 1 ) + f ( x n ) = A n n f ( x 1 ) + f ( x 2 ) + · · · + f ( x n − 1 ) + f ( x n ) · b − a = n b − a b − a · [ f ( x 1 ) + f ( x 2 ) + · · · + f ( x n − 1 ) + f ( x n )] · b − a 1 = n 1 = b − a · [ f ( x 1 )∆ x + f ( x 2 )∆ x + · · · + f ( x n − 1 )∆ x + f ( x n )∆ x ] 1 = b − a · R n , � b where R n is the n th right Riemann sum for the integral a f ( x ) dx . D.L. White (Kent State University) 4 / 7
Average Value of a Function 1 Finally, since b − a is a constant, we have that the average value f ave of f ( x ) on the interval [ a , b ] is � b 1 1 1 f ave = lim n →∞ A n = lim b − aR n = b − a lim n →∞ R n = f ( x ) dx . b − a n →∞ a Average Value If y = f ( x ) is a continuous function, then the AVERAGE VALUE of f on the interval [ a , b ] is � b 1 f ave = f ( x ) dx . b − a a D.L. White (Kent State University) 5 / 7
Example 1 Find the average value of f ( x ) = x 2 + 4 on the interval [3 , 12]. The average value is � 12 � 12 � � 1 1 1 ( x 2 + 4) dx 3 x 3 + 4 x � = � 12 − 3 9 � 3 3 1 �� 1 � � 1 �� 3(12 3 ) + 4(12) 3(3 3 ) + 4(3) = − 9 1 = 9 [(576 + 48) − (9 + 12)] 1 9 [624 − 21] = 603 = = 67 . 9 � 12 3 ( x 2 + 4) dx = 603 is NOT the average value. Note that The integral must be divided by the length of the interval. D.L. White (Kent State University) 6 / 7
Average Velocity Finally, suppose an object is moving along a straight line so that its position at time t is s ( t ), and so its velocity at time t is v ( t ) = s ′ ( t ). Previously, we discussed average velocity on a time interval t = a to t = b . We now also have the average value of the velocity function on [ a , b ]. Are they the same thing? Recall that average velocity on the interval t = a to t = b is ∆ s ∆ t = s ( b ) − s ( a ) . b − a Since s ( t ) is an antiderivative for v ( t ), we have that the average value of the velocity function on [ a , b ] is � b b 1 1 � b − a [ s ( b ) − s ( a )] = s ( b ) − s ( a ) 1 � v ( t ) dt = b − a · s ( t ) = . � b − a b − a � a a Thus average velocity is equal to the average value of the velocity function. D.L. White (Kent State University) 7 / 7
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