Question

Calcule a integral por partes

Answer 1

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


Método de integração por partes:

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


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \:x\cdot \:2^{-x}dx\\\\\\=-\frac{2^{-x}x}{ln \left(2\right)}-\int \:-\frac{2^{-x}}{ln \left(2\right)}dx\\\\\\=-\frac{2^{-x}x}{ln \left(2\right)}-\frac{1}{ln ^2\left(2\right)}\cdot \:2^{-x}\\\\\\\to \boxed{\sf =-\frac{2^{-x}x}{ln \left(2\right)}-\frac{1}{ln ^2\left(2\right)}\cdot \:2^{-x}+C}$