Ramiro Wise

Answered

2022-11-28

Find the average value fave of the function f on the given interval.

$f(t)=e\mathrm{sin}(t),\mathrm{cos}(t),[0,\frac{\pi}{2}]$

$f(t)=e\mathrm{sin}(t),\mathrm{cos}(t),[0,\frac{\pi}{2}]$

Answer & Explanation

rezvanifanFdT

Expert

2022-11-29Added 9 answers

The average value of a function over an interval $[a,b]$ is given by

$\frac{1}{b-a}{\int}_{a}^{b}f(x)dx$

We have

$\frac{1}{\frac{\pi}{2}-0}{\int}_{0}^{\pi /2}{e}^{\mathrm{sin}t}\mathrm{cos}tdt=\frac{2}{\pi}{\int}_{0}^{\pi /2}{e}^{\mathrm{sin}t}d(\mathrm{sin}t)\phantom{\rule{0ex}{0ex}}=\frac{2}{\pi}{\int}_{\mathrm{sin}0}^{\mathrm{sin}(\pi /2)}{e}^{u}du=\frac{2}{\pi}{e}^{u}{|}_{\mathrm{sin}0}^{\mathrm{sin}(\pi /2)}\phantom{\rule{0ex}{0ex}}=\frac{2({e}^{1}-{e}^{0})}{\pi}=\frac{2(e-1)}{\pi}$

$\frac{1}{b-a}{\int}_{a}^{b}f(x)dx$

We have

$\frac{1}{\frac{\pi}{2}-0}{\int}_{0}^{\pi /2}{e}^{\mathrm{sin}t}\mathrm{cos}tdt=\frac{2}{\pi}{\int}_{0}^{\pi /2}{e}^{\mathrm{sin}t}d(\mathrm{sin}t)\phantom{\rule{0ex}{0ex}}=\frac{2}{\pi}{\int}_{\mathrm{sin}0}^{\mathrm{sin}(\pi /2)}{e}^{u}du=\frac{2}{\pi}{e}^{u}{|}_{\mathrm{sin}0}^{\mathrm{sin}(\pi /2)}\phantom{\rule{0ex}{0ex}}=\frac{2({e}^{1}-{e}^{0})}{\pi}=\frac{2(e-1)}{\pi}$

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