f ( x ) = sin ( 4 x ) , [ − π , π ] {\displaystyle f(x)=\sin {(4x)}{\text{,}}\quad [-\pi ,\pi ]}
f a v g = 1 π − ( − π ) ∫ − π π sin ( 4 x ) d x = 1 2 π ∫ − π π sin ( 4 x ) d x = 1 2 π ∫ − 4 π 4 π sin ( u ) ( 1 4 d u ) = 1 8 π ∫ − 4 π 4 π sin ( u ) d u = − 1 8 π cos ( u ) | − 4 π 4 π = [ − 1 8 π cos ( 4 π ) ] − [ − 1 8 π cos ( − 4 π ) ] = [ − 1 8 π ( 1 ) ] + [ 1 8 π ( 1 ) ] = 0 {\displaystyle {\begin{aligned}f_{avg}&={\frac {1}{\pi -(-\pi )}}\int _{-\pi }^{\pi }\sin {(4x)}\,dx={\frac {1}{2\pi }}\int _{-\pi }^{\pi }\sin {(4x)}\,dx\\[2ex]&={\frac {1}{2\pi }}\int _{-4\pi }^{4\pi }\sin {(u)}\left({\frac {1}{4}}\,du\right)={\frac {1}{8\pi }}\int _{-4\pi }^{4\pi }\sin(u)\,du\\[2ex]&=-{\frac {1}{8\pi }}\cos(u){\bigg |}_{-4\pi }^{4\pi }\\[2ex]&=\left[-{\frac {1}{8\pi }}\cos(4\pi )\right]-\left[-{\frac {1}{8\pi }}\cos(-4\pi )\right]=\left[-{\frac {1}{8\pi }}(1)\right]+\left[{\frac {1}{8\pi }}(1)\right]\\[2ex]&=0\end{aligned}}}
u = 4 x d u = 4 d x 1 4 d u = d x {\displaystyle {\begin{aligned}u&=4x\\[2ex]du&=4dx\\[2ex]{\frac {1}{4}}du&=dx\end{aligned}}}
New upper limit: 4 π = 4 ( π ) {\displaystyle 4\pi =4(\pi )} New lower limit: − 4 π = 4 ( − π ) {\displaystyle -4\pi =4(-\pi )}
![{\displaystyle f(x)=\sin {(4x)}{\text{,}}\quad [-\pi ,\pi ]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0d26264e15aeb29ff049f70b6f38de6ed8709857)
![{\displaystyle {\begin{aligned}f_{avg}&={\frac {1}{\pi -(-\pi )}}\int _{-\pi }^{\pi }\sin {(4x)}\,dx={\frac {1}{2\pi }}\int _{-\pi }^{\pi }\sin {(4x)}\,dx\\[2ex]&={\frac {1}{2\pi }}\int _{-4\pi }^{4\pi }\sin {(u)}\left({\frac {1}{4}}\,du\right)={\frac {1}{8\pi }}\int _{-4\pi }^{4\pi }\sin(u)\,du\\[2ex]&=-{\frac {1}{8\pi }}\cos(u){\bigg |}_{-4\pi }^{4\pi }\\[2ex]&=\left[-{\frac {1}{8\pi }}\cos(4\pi )\right]-\left[-{\frac {1}{8\pi }}\cos(-4\pi )\right]=\left[-{\frac {1}{8\pi }}(1)\right]+\left[{\frac {1}{8\pi }}(1)\right]\\[2ex]&=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d09e47433b09ec847b4ea554ffe72cb9c71f26c1)
![{\displaystyle {\begin{aligned}u&=4x\\[2ex]du&=4dx\\[2ex]{\frac {1}{4}}du&=dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/2d7068b9cc23779b7a691ebbcdd78a156d8eab11)
