Its volume is r r r π r 2 x 2 2 dx π r r r r 2 x 2

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Its volume is R r - r π ( r 2 - x 2 ) 2 dx = π R r - r ( r 2 - x 2 ) dx = π ( r 2 x - 1 3 x 3 ) | r - r = π ( 2 3 r 3 - - 2 3 r 3 ) = 4 3 π r 3 . 47
Integration techniques 8.2: Volumes of revolution and average value Average value Definition If y 1 , . . . , y n is a finite sequence of real numbers, then the average or mean value of the sequence is μ = y 1 + · · · + y n n What would be meant by the average value of a function over some interval? For example, what is the average value of f ( x ) = x 2 over the interval from 1 to 3? 48
Integration techniques 8.2: Volumes of revolution and average value Average value, imprecisely If f ( x ) is a continuous function defined on some interval [ a , b ], we can define the average value of f ( x ) from the notion of the average of a finite sequence. Fix a number n and sample consider the sequence f ( x 1 ) , . . . , f ( x n ) where a x 1 x 2 ≤ · · · ≤ x n b is a sequence of points between [ a , b ] chosen with a “uniform” distribution. 49
Integration techniques 8.2: Volumes of revolution and average value Average value, precisely To make the notion of “uniform” precise, we could break [ a , b ] into n intervals each of length Δ x = b - a n and choose x i in the i th interval. The average of the sequence is then μ = f ( x 1 ) + · · · + f ( x n ) n = ( f ( x 1 ) + · · · + f ( x n )) 1 n = 1 b - a ( f ( x 1 ) + · · · + f ( x n )) b - a n = 1 b - a ( f ( x 1 ) + · · · + f ( x n ))Δ x 1 b - a [Riemann sum of f ( x ) with respect to n and x 1 , . . . , x n ] 50
Integration techniques 8.2: Volumes of revolution and average value Definition of average value As the Riemann sums approach the integral R b a f ( x ) dx , we may define the average value of a continuous function over an interval in terms of its integral. Definition Let f ( x ) be a continuous function on the interval [ a , b ] we define the average value or mean of f ( x ) over [ a , b ] to be μ = 1 b - a Z b a f ( x ) dx 51
Integration techniques 8.2: Volumes of revolution and average value The example of f ( x ) = x 2 We compute that the average value of f ( x ) = x 2 over the interval from x = 1 to x = 3 is μ = 1 3 - 1 Z 3 1 x 2 dx = 1 2 Z 3 1 x 2 dx = 1 2 ( 1 3 x 3 ) | 3 1 = 1 6 (27 - 1) = 13 3 = 4 1 3 52
Integration techniques 8.2: Volumes of revolution and average value An average with a more difficult integral What is the average value of f ( x ) = 3 x + 1 + xe - x 2 over the interval - 1 to 1? 53
Integration techniques 8.2: Volumes of revolution and average value Solution The average is μ = 1 1 - ( - 1) Z 1 - 1 (3 x + 1 + xe - x 2 ) dx = 1 2 (3 Z 1 - 1 xdx + Z 1 - 1 dx + Z 1 - 1 xe - x 2 dx ) = 1 2 ( 3 2 x 2 | 1 - 1 + x | 1 - 1 + Z 1 - 1 xe - x 2 dx ) = 1 2 (0 + 2 + Z 1 - 1 xe - x 2 dx ) = 1 + 1 2 Z 1 - 1 xe - x 2 dx 54
Integration techniques 8.2: Volumes of revolution and average value Computation of 1 2 R 1 - 1 xe - x 2 dx If g ( x ) = xe - x 2 , then g ( - x ) = ( - x ) e - ( - x ) 2 = - xe - x 2 = - g ( x ). Thus, R 0 - 1 xe - x 2 dx = - R 1 0 xe - x 2 dx and 1 2 R 1 - 1 xe - x 2 dx = 0 yielding μ = 1. 55
Integration techniques 8.2: Volumes of revolution and average value A symbolic solution Take u = - x 2 so that du = - 2 xdx to compute Z xe - x 2 dx = Z - 1 2 - 2 xe - x 2 dx = - 1 2 Z e - x 2 - 2 xdx = - 1 2 Z e u du = - 1 2 e u + C = - 1 2 e - x 2 + C Hence, Z 1 - 1 xe - x 2 dx = - 1 2 e - 1 - - 1 2 e - 1 = 0 .