What is the average value of the function $f (x) = x - (x^{2})$ $f(x) = x - (x^2)$ on the interval $[0, 2]$ $[0,2]$ ?

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What is the average value of the function $f (x) = x - (x^{2})$ $f(x) = x - (x^2)$ on the interval $[0, 2]$ $[0,2]$ ?

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Answer 1 · 2016-09-21T17:54:20

The average value of $f$ $f$ on $[a, b}$ $[a,b}$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ $1/(b-a) int_a^b f(x) dx$ .

Explanation:

For this function on this interval, I get $- \frac{1}{3}$ $-1/3$

Ave $= \frac{1}{2 - 0} \int_{0}^{2} (x - x^{2}) d x$ $= 1/(2-0) int_0^2 (x-x^2) dx$

$= \frac{1}{2} {[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{2}$ $= 1/2[x^2/2-x^3/3]_0^2$

$= \frac{1}{2} [(\frac{4}{2} - \frac{8}{3}) - (0)]$ $= 1/2[(4/2-8/3)-(0)]$

$= \frac{1}{2} (- \frac{2}{3}) = - \frac{1}{3}$ $= 1/2(-2/3) = -1/3$

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What is the average value of the function $f (x) = x - (x^{2})$ $f(x) = x - (x^2)$ on the interval $[0, 2]$ $[0,2]$ ?

1 Answer

Explanation:

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