6.5 Average Value of a Function/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 f(t) = te^{-t^2} \quad [0, 5] }

${\begin{aligned}f_{avg}&={\frac {1}{5}}\int _{0}^{5}te^{-t^{2}}\,dx\\[2ex]&={\frac {1}{5}}\int _{0}^{5}-{\frac {1}{2}}(e^{u})\,du\\[2ex]&={\frac {1}{5}}\int _{0}^{25}e^{u}(-{\frac {1}{2}}du)\\[2ex]&={\frac {1}{10}}\int _{-25}^{0}e^{u}\,du\\[2ex]&={\frac {1}{10}}e^{u}{\bigg |}_{-25}^{0}\\[2ex]&={\frac {1}{10}}-{\frac {1}{10}}e^{-25}\\[2ex]&={\frac {1}{10}}(1-e^{-25})\\[2ex]\end{aligned}}$

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} & u=t^2 \\[2ex] & \frac{du}{dt}=2t \\[2ex] & du=-2t\cdot{dt} \\[2ex] & -\frac{1}{2}du=t\cdot{dt} \\[2ex] \end{align} }