Question

Find the average value of the function on the given interval\({\rm{h(x) = co}}{{\rm{s}}^{\rm{4}}}{\rm{xsinx,}}\;\;\;{\rm{(0,\pi )}}{\rm{.}}\)

Short answer

The average value of the function interval is \({{\rm{h}}_{{\rm{avg}}}}{\rm{ = }}\frac{{\rm{2}}}{{{\rm{5\pi }}}}\).

Step-by-step solution

Step 1:Formula used.

The average value of the function \({\rm{f}}\left( {\rm{x}} \right)\)over the interval \(\left( {{\rm{a, b}}} \right)\)is given by

\(\) \({{\rm{f}}_{{\rm{avg}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{b - a}}}}\int_{\rm{a}}^{\rm{b}} {\rm{f}} {\rm{(x)dx}}\)

The average value of\({\rm{h(x) = co}}{{\rm{s}}^{\rm{4}}}\)\({\rm{x sin x}}\)over the interval \({\rm{(0,\pi )}}\) is given by

\(\) \({{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi - 0}}}}\int_{\rm{0}}^{\rm{\pi }} {{\rm{co}}{{\rm{s}}^{\rm{4}}}} {\rm{ x sin x dx}}\)

\(\) \({{\rm{h}}_{{\rm{avg}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}\int_{\rm{0}}^{\rm{\pi }} {{\rm{co}}{{\rm{s}}^{\rm{4}}}} {\rm{ x(sin x dx)}}{\rm{.}}\)

Calculation.

Substitute \({\rm{cos x = u and - sin x dx = du}}\)

Limits of Integration will change from \(int_0^\pi {\rm{ to }}\int_{\cos (0)}^{\cos (\pi )} = \mathop \smallint \nolimits_1^{ - 1} \)

\(\begin{aligned}{c}{{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}\int_{\rm{1}}^{{\rm{ - 1}}} {{{\rm{u}}^{\rm{4}}}} {\rm{( - du)}}\\{{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}\int_{\rm{1}}^{{\rm{ - 1}}} {\rm{ - }} {{\rm{u}}^{\rm{4}}}{\rm{du}}\\{{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}\left( {{\rm{ - }}\frac{{{{\rm{u}}^{{\rm{4 + 1}}}}}}{{{\rm{4 + 1}}}}} \right)_{\rm{1}}^{{\rm{ - 1}}}\end{aligned}\)

\({{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}\left( {{\rm{ - }}\frac{{{{\rm{u}}^{\rm{5}}}}}{{\rm{5}}}} \right)_{\rm{1}}^{{\rm{ - 1}}}\)

\(\begin{aligned}{c}{{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}\left( {{\rm{ - }}\frac{{{{{\rm{( - 1)}}}^{\rm{5}}}}}{{\rm{5}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{\rm{\pi }}}\left( {{\rm{ - }}\frac{{{{{\rm{(1)}}}^{\rm{5}}}}}{{\rm{5}}}} \right)\\{{\rm{h}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{5\pi }}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{5\pi }}}}\end{aligned}\)

Hence the average value of the function interval \({{\rm{h}}_{{\rm{avg}}}}{\rm{ = }}\frac{{\rm{2}}}{{{\rm{5\pi }}}}.\)