Answer with Step-by-step explanation:
We are given that
[tex]f(x)=-sin x[/tex]
[tex][0,\infty][/tex]]
Average value of the function is given gy
[tex]f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{\pi-0}\int_{0}^{\pi}-sinx dx[/tex]
[tex]f_{avg}=\frac{1}{\pi}[cosx]^{\pi}_{0}[/tex]
Using the formula
[tex]\int sin xdx=-cos x[/tex]
[tex]f_{avg}=\frac{1}{\pi}(cos\pi-cos0)[/tex]
[tex]f_{avg}=\frac{1}{\pi}(-1-1)=-\frac{2}{\pi}[/tex]
[tex]f(x)=f_{avg}[/tex]
[tex]-sinx=-\frac{2}{\pi}[/tex]
[tex]sinx=\frac{2}{\pi}[/tex]
[tex]x=sin^{-1}(\frac{2}{\pi})=0.69radian[/tex]
The average value of the function is [tex]-\frac{2}{\pi}[/tex].
The average value of the function is given as,
[tex]Average=\frac{1}{\pi} \int\limits^\pi_0 {f(x)} \, dx[/tex]
Given function is, [tex]f(x)=-sinx[/tex]
Substitute values in above relation.
[tex]Average=\frac{1}{\pi} \int\limits^\pi_0 {-sinx} \, dx\\\\Average=\frac{1}{\pi} (cosx)^{\pi} _{0}\\\\Average=\frac{1}{\pi}(cos\pi - cos 0)\\\\Average=\frac{1}{\pi}(-1-1)\\\\Average=-\frac{2}{\pi}[/tex]
The values of x in the interval for which the function equals its average value is,
[tex]-sinx=-\frac{2}{\pi}\\ \\x=sin^{-1} (\frac{2}{\pi} )=39.56[/tex]
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