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Math 104 – Rimmer 6.5 Average Value of a Function Average of n numbers: x1 + x2 + x3 + xn n Take the n numbers to be sample points for a function: ( ) ( ) ( ) ( ) f x1* + f x2* + f x3* + f xn* n Partition the interval [ a, b ] into n subintervals of equal length. Here n = 8. a b What is the width of each subinterval? We now have: b−a n ∆x = n ( ) ( ) ( ) = ∑ f ( x ) ∆x ( ) b−a b−a ⇒n= n ∆x * f x1* + f x2* + f x3* + f xn* b−a ∆x Call this ∆x i i =1 b−a n Math 104 – Rimmer 6.5 Average Value of a Function lim ∑ f xi ∆x ( ) * Taking the limit as n → ∞, we get n →∞ i =1 b−a b 1 f ( x ) dx Average value of f ( x ) on an interval [ a, b ] = b − a ∫a Find the average value of the function Average value 6 1 3 3 dr = ∫ 2 of f ( x ) = on 1, 6 [ ] 5 2 1 (1 + r ) (1 + r ) = 3 (1 + r ) 2 on the interval [1, 6] . u =1+ r r = 1⇒ u = 2 du = dr r = 6 ⇒ u = 7 7 7 7 1 3 3 −2 3 −1 = 3 − 1 − − 1 du = u du = 2 5 ∫2 u 2 5 ∫2 5 u 2 5 7 3 −2 7 3 5 3 = + = = 5 14 14 5 14 14 1 Math 104 – Rimmer 6.5 Average Value of a Function Let f be continuous on [ a, b ] , then there is a value c in [ a, b ] such that b 1 f (c) = f ( x ) dx b − a ∫a or b f ( c )( b − a ) = ∫ f ( x ) dx a area of the rectangle area under the curve = Math 104 – Rimmer 6.5 Average Value of a Function Proof: Let f be continuous on [ a, b ] , f ( x) By the Extreme Value Theorem, there is a m and M such that m ≤ f ( x ) ≤ M . b Then m ( b − a ) ≤ ∫ f ( x ) dx ≤ M ( b − a ) a b ⇒ m≤ 1 f ( x ) dx ≤ M b − a ∫a By the Intermediate Value Theorem, b there is a c in [ a, b ] with f ( c ) = 1 f ( x ) dx. b − a ∫a 2 Math 104 – Rimmer 6.5 Average Value of a Function (a) Find the average value of f ( x ) = x on the interval [0, 4]. (b) Find c such that f ave = f ( c ) . (c) Sketch the graph of f and a rectangle whose area is the same as the area under the graph of f . Average value ( a ) of f ( x) on [ a, b ] b = 4 Average value 1 1 1/ 2 f x dx ⇒ ( ) x dx = b − a ∫a of f ( x ) = x on [ 0, 4] 4 ∫0 4 = 4 4 16 (b ) f ( c ) = ⇒ c = ⇒ c = 3 3 9 = 1 1 2 3/ 2 x = 43/ 2 4 3 0 6 ( 1 1/ 2 4 6 3 ( ) = 8 = 6 ) 4 3 (c) 3