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

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Answer 1 · 2016-10-27T16:06:21

$98$ $98$

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

$\frac{1}{b - a} \int_{a}^{b} f (x) d x$ $1/(b-a) int_a^b f(x) dx$ .

For this problem, that is

$\frac{1}{10 - 0} \int_{0}^{10} (18 x + 8) d x = \frac{1}{10} {[9 x^{2} + 8 x]}_{0}^{2}$ $1/(10-0) int_0^10 (18x+8) dx = 1/10 [9x^2+8x]_0^10$

$= \frac{1}{10} [980] = 98$ $= 1/10[980] = 98$ .

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