7.1 Integration By Parts/25

$\int _{0}^{1}{\frac {y}{e^{2y}}}dy$

$u=y\qquad du=dy$

$dv={\frac {dy}{e^{2}y}}=e^{-2y}dy\qquad v={\frac {-1}{2}}e^{-2y}$

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