6.5 Average Value of a Function/5

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

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