6.5 Average Value of a Function/17

1. Use the Average Value from a to b:

$f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx$

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{12}}\int _{0}^{12}50+\underbrace {14\sin({\frac {\pi }{12}}t)} \,dt={\frac {1}{12}}[50t-{\frac {168}{\pi }}\cos({\frac {\pi }{12}}t)]{\bigg |}_{12}^{0}={\frac {1}{12}}[(50)(12)-{\frac {168}{\pi }}\cos(\pi ))(0-{\frac {168}{\pi }}\cos(0)]={\frac {1}{12}}[600-{\frac {168}{\pi }}(-1)+{\frac {168}{\pi }}(1)]={\frac {1}{12}}[600+{\frac {168}{\pi }}+{\frac {168}{\pi }}]}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{aligned} u &= \frac{\pi}{12}t \\ dt\cdot\frac{du}{dt} &= dt \\ \frac{12}{\pi}du &= dt \\ integrate for 14\sin(u)\frac{12}{\pi} \\ \int14\sin(u)\frac{12}{\pi}\,du \\ 14\cdot\frac{12}{\pi}\int\sin(u)\,du \\ -\frac{168}{\pi}\cos(u) \\ -\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\ \end{aligned} }