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] | ||
</math> | |||
<math> | |||
\begin{align} | \begin{align} | ||
f_{avg} = \frac{1}{5}\int_{0}^{5}te^{-t^2}\,dx \\[2ex] | f_{avg} = \frac{1}{5}\int_{0}^{5}te^{-t^2}\,dx \\[2ex] | ||
Revision as of 19:01, 16 December 2022
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 f(t) = te^{-t^2} \quad [0, 5] }
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{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} }