7.1 Integration By Parts/25

${\begin{aligned}\int _{0}^{1}{\frac {y}{e^{2y}}}dy\\[2ex]&u=y,-du=dy,\\[2ex]&dv={\frac {dy}{e^{2}y}}=e^{-2y}dy,v={\frac {-1}{2}}e^{-2y}\end{aligned}}$

${\begin{aligned}\int _{0}^{1}{\frac {y}{e^{2y}}}dy&=\left[{\frac {-1}{2}}ye^{-2y}\right]{\bigg |}_{0}^{1}+{\frac {1}{2}}\int _{0}{1}e^{-2y}dy\\[2ex]&=\left({\frac {-1}{2}}e^{-2}+0\right)-{\frac {1}{4}}\left[e^{-2y}\right]{\bigg |}_{0}^{1}\\[2ex]&={\frac {-1}{2}}e^{-2}-{\frac {1}{4}}e^{-2}+{\frac {1}{4}}\\[2ex]&={\frac {1}{4}}-{\frac {3}{4}}e^{-2}\\[2ex]\end{aligned}}$