Question

Find the average value of the function on the given interval

\(\) \({\rm{f(x) = sin4x,}}\;\;\;{\rm{( - \pi ,\pi )}}{\rm{.}}\)

Short answer

The average value of the function interval is \({\rm{o}}\).

Step-by-step solution

Step 1:Formula used.

\(f(x) = \sin (4x)\)

On \({\rm{( - \pi ,\pi )}}\)

Using this formula, calculate the average value of this function.

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

Where \({\rm{a}}\) and \({\rm{b}}\) are domain endpoints.

Calculation.

\(\)\({{\rm{f}}_{{\rm{avg }}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi - ( - \pi )}}}}\int_{{\rm{ - \pi }}}^{\rm{\pi }} {{\rm{sin}}} {\rm{(4x)dx}}\)

\(\) \( = \frac{1}{{2\pi }} \cdot \int_{ - \pi }^\pi {\sin } (4x){\rm{d}}x = \left( {\begin{aligned}{*{20}{c}}{t = 4x\;{\rm{d}}t = 4\;{\rm{d}}x \to {\rm{d}}x = \frac{{{\rm{d}}t}}{4}}\end{aligned}} \right)\)

\( = \frac{1}{{2\pi }} \cdot \int_{ - 4\pi }^{4\pi } {\sin } t \cdot \frac{{{\rm{d}}t}}{4}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{2\pi }}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\int_{{\rm{ - 4\pi }}}^{{\rm{4\pi }}} {{\rm{sin}}} {\rm{t\;dt}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{8\pi }}}}{\rm{ \times }}\left( {{\rm{ - }}\left. {{\rm{cosu}}} \right|_{{\rm{ - 4\pi }}}^{{\rm{4\pi }}}} \right)\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{8\pi }}}}{\rm{ \times }}\left( {{\rm{ - }}\left. {{\rm{cos(4x)}}} \right|_{{\rm{ - \pi }}}^{\rm{\pi }}} \right)\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{8\pi }}}}{\rm{ \times ( - cos(4\pi ) - ( - cos( - 4\pi ))}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{8\pi }}}}{\rm{ \times 0}}\)

Hence the average value of the function interval = \({\rm{0}}{\rm{.}}\)