NormmodulxEE

2022-12-03

Find the average value ${f}_{ave}$ of the function f on the given interval.
$f\left(x\right)=3{x}^{2}+8x,\left[-1,2\right]$

revessatrP6

Expert

$f\left(x\right)=3{x}^{2}+8x,\left[-1,2\right]$
Average value of $f\left(x\right)=\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2-\left(-1\right)}{\int }_{-1}^{2}\left(3{x}^{2}+8x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[\frac{3{x}^{3}}{3}+\frac{8{x}^{2}}{2}{\right]}_{-1}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left({x}^{3}+4{x}^{2}{\right)}_{-1}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{2}^{3}+4×{2}^{2}-\left(\left(-1{\right)}^{3}+4×\left(-1{\right)}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[8+16-\left(\left(-1\right)+4\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[24-3\right]=\frac{1}{3}×21=7$

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