y = x e − 0.4 x x = 0 , x = 1 {\displaystyle y=xe^{-0.4x}~~~x=0,x=1}
∫ 0 5 x e − 0.4 x d x ⏟ u = x d v = e − 0.4 x d x d u = d x v = 5 2 e − 0.4 x {\displaystyle {\begin{aligned}\underbrace {\int _{0}^{5}xe^{-0.4x}dx} _{\begin{aligned}u&=x&dv&=e^{-0.4x}dx\\[0.6ex]du&=dx&v&={\frac {5}{2}}e^{-0.4x}\end{aligned}}\\[2ex]\end{aligned}}}
[ 5 2 e − 0.4 x ] | 0 1 + ∫ 5 0 5 2 e − 0.4 x d x {\displaystyle \left[{\frac {5}{2}}e^{-0.4x}\right]{\bigg |}_{0}^{1}+\int _{5}^{0}{\frac {5}{2}}e^{-0.4x}dx}
u = − 0.4 x d u = − 0.4 d x {\displaystyle {\begin{aligned}u&=-0.4x\\[2ex]du&=-0.4dx\\[2ex]\end{aligned}}}
( − 25 2 e − 2 − 0 ) + 5 4 ∫ 5 0 e u {\displaystyle \left(-{\frac {25}{2}}e^{-2}-0\right)+{\frac {5}{4}}\int _{5}^{0}e^{u}}
− 25 2 e − 2 + [ 5 4 ( e − 0.4 x ) ] = − 25 2 e − 2 − 25 4 e − 2 + 25 4 = − 75 4 e − 2 + 25 4 {\displaystyle -{\frac {25}{2}}e^{-2}+\left[{\frac {5}{4}}\left(e^{-0.4x}\right)\right]~~~=~~~-{\frac {25}{2}}e^{-2}-{\frac {25}{4}}e^{-2}+{\frac {25}{4}}~~~=~~~-{\frac {75}{4}}e^{-2}+{\frac {25}{4}}}