Keenan Santos

2022-07-05

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

Pranav Greer

Beginner2022-07-06Added 13 answers

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

Ave $=\frac{1}{2-0}{\int}_{0}^{2}(x-{x}^{2})dx$

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

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

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

Ave $=\frac{1}{2-0}{\int}_{0}^{2}(x-{x}^{2})dx$

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

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

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

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