6.5 Average Value of a Function/5: Difference between revisions
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\begin{align} | \begin{align} | ||
f_{avg} = \frac{1}{5}\int_{0}^{5}te^{-t^2}\,dx = \frac{1}{5}\int_{0}^{5}-\frac{1}{2}(e^u)\,du = \frac{1}{5}\int_{0}^{25}e^u(-\frac{1}{2}du) = \frac{1}{10}\int_{-25}^{0}e^u\,du = \frac{1}{10}e^u \bigg|_{-25}^{0} = \frac{1}{10}-\frac{1}{10}e^{-25} = \frac{1}{10}(1-e^{-25}) | 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{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 18:59, 16 December 2022
Failed to parse (syntax error): {\displaystyle f(t) = te^{-t^2} \quad [0, 5] \\ \begin{align} 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{align} }