Question

integral de x*2^-x integral por parte

Answer 1

$I=\displaystyle\int{x\cdot 2^{-x}\,dx}$


Método de integração por partes:

$\begin{array}{lcl} u=x&~\Rightarrow~&du=1\,dx\\\\ dv=2^{-x}\,dx&~\Leftarrow~&v=-\dfrac{2^{-x}}{\mathrm{\ell n\,}2} \end{array}\\\\\\\\ \displaystyle\int{u\,dv}=uv-\int{v\,du}\\\\\\ \int{x\cdot 2^{-x}\,dx}=-\dfrac{x\cdot 2^{-x}}{\mathrm{\ell n\,}2}-\int{\left(-\dfrac{2^{-x}}{\mathrm{\ell n\,}2} \right )\cdot 1\,dx}\\\\\\ I=-\dfrac{x\cdot 2^{-x}}{\mathrm{\ell n\,}2}+\dfrac{1}{\mathrm{\ell n\,}2}\int{2^{-x}\,dx}\\\\\\ I=-\dfrac{x\cdot 2^{-x}}{\mathrm{\ell n\,}2}+\dfrac{1}{\mathrm{\ell n\,}2}\cdot \left(-\dfrac{2^{-x}}{\mathrm{\ell n\,}2} \right )+C\\\\\\ I=-\dfrac{x\cdot 2^{-x}}{\mathrm{\ell n\,}2}-\dfrac{2^{-x}}{(\mathrm{\ell n\,}2)^{2}}+C$

$\therefore~~\boxed{\begin{array}{c}\displaystyle\int{x\cdot 2^{-x}\,dx}=-\dfrac{x\cdot 2^{-x}}{\mathrm{\ell n\,}2}-\dfrac{2^{-x}}{(\mathrm{\ell n\,}2)^{2}}+C \end{array}}$