Question

solve integral of y*e^-y dy=​

Answer 1

Solution:

Given integral:

$\displaystyle \tt \longrightarrow I = \int y {e}^{ - y} \: dy$

We will solve this problem by integral by parts.

Let us assume that:

$\tt \longrightarrow u =y \: or\: du =dy$

$\tt \longrightarrow dv = {e}^{ - y} \: dy \: or\:v = - {e}^{ - y}$

So, the integral changes to:

$\displaystyle \tt \longrightarrow I = uv - \int v \: du$

$\displaystyle \tt \longrightarrow I = - y {e}^{ - y} - \int - {e}^{ - y} \: dy$

$\displaystyle \tt \longrightarrow I = - y {e}^{ - y} + \int{e}^{ - y} \: dy$

$\displaystyle \tt \longrightarrow I = - y {e}^{ - y} - {e}^{ - y} + C$

$\displaystyle \tt \longrightarrow I = \frac{ - y}{ {e}^{ - y} } - \dfrac{1}{ {e}^{ - y}} + C$

$\displaystyle \tt \longrightarrow I = \frac{ - (y + 1)}{ {e}^{ - y} } + C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$