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What is the average value of the function $f(x)=18x+8$ on the interval $[0,10]$ ?

1 Answer

Oct 27, 2016

$98$

Explanation:

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

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

For this problem, that is

$1/(10-0) int_0^10 (18x+8) dx = 1/10 [9x^2+8x]_0^10$

$= 1/10[980] = 98$ .

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