f ( t ) = t e − t 2 [ 0 , 5 ] f a v g = 1 5 ∫ 0 5 t e − t 2 d x = 1 5 ∫ 0 5 − 1 2 ( e u ) d u = 1 5 ∫ 0 25 e u ( − 1 2 d u ) = 1 10 ∫ − 25 0 e u d u = 1 10 e u | − 25 0 = 1 10 − 1 10 e − 25 = 1 10 ( 1 − e − 25 ) {\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}}}
![{\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cfe05c15b90fcf92be287628cd5382a650d1936a)