6.5 Average Value of a Function/5: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
f_{avg} = \frac{1}{5}\int_{0}^{5}te^{-t^2}\,dx | f_{avg} &= \frac{1}{5}\int_{0}^{5}te^{-t^2}\,dx = \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) = \frac{1}{10}\int_{-25}^{0}e^u\,du \\[2ex] | |||
& = \frac{1}{5}\int_{0}^{25}e^u(-\frac{1}{2}du) | |||
&= \frac{1}{10}e^u \bigg|_{-25}^{0} = \frac{1}{10}-\frac{1}{10}e^{-25} = \frac{1}{10}(1-e^{-25}) \\[2ex] | |||
& = \frac{1}{10}e^u \bigg|_{-25}^{0} | \end{align} | ||
</math> | |||
<math> | |||
\begin{align} | |||
& u=t^2 \\[2ex] | & u=t^2 \\[2ex] | ||
& \frac{du}{dt}=2t \\[2ex] | & \frac{du}{dt}=2t \\[2ex] |
Latest revision as of 19:46, 16 December 2022