6.5 Average Value of a Function/5: Difference between revisions

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<math>
<math>
f(t) = te^{-t^2} \quad [0, 5] \\
f(t) = te^{-t^2} \quad [0, 5]


\begin{align}
\begin{align}

Revision as of 19:00, 16 December 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(t)=te^{-t^{2}}\quad [0,5]{\begin{aligned}f_{avg}={\frac {1}{5}}\int _{0}^{5}te^{-t^{2}}\,dx\\[2ex]&={\frac {1}{5}}\int _{0}^{5}-{\frac {1}{2}}(e^{u})\,du\\[2ex]&={\frac {1}{5}}\int _{0}^{25}e^{u}(-{\frac {1}{2}}du)\\[2ex]&={\frac {1}{10}}\int _{-25}^{0}e^{u}\,du\\[2ex]&={\frac {1}{10}}e^{u}{\bigg |}_{-25}^{0}\\[2ex]&={\frac {1}{10}}-{\frac {1}{10}}e^{-25}\\[2ex]&={\frac {1}{10}}(1-e^{-25})\\[2ex]\end{aligned}}}