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What is the average value of the function $f(x)=(x-1)^2$ on the interval from x=1 to x=5?

1 Answer

Mar 24, 2016

The average value is $16/3$

Explanation:

The average value of a function $f$ on an interval $[a,b]$ is

$1/(b-a) int_a^b f(x) dx$

So the value we seek is

$1/(5-1) int_1^5 (x-1)^2 dx = 1/4[(x-1)^3/3]_1^5$

$= 1/12[(4)^3-(0)^3]$

$= 16/3$

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