Question

estou com dificuldade para resolver uma integral dupla 5a0 e 1a0 da função (y+xe^y)dydx

Answer 1

$\displaystyle\int\limits_{0}^{5}\int\limits_{0}^{1}{(y+xe^{y})\,dy\,dx}\\ \\ \\ =\int\limits_{0}^{5}\int\limits_{0}^{1}{y\,dy\,dx}+\int\limits_{0}^{5}\int\limits_{0}^{1}{xe^{y}\,dy\,dx}$


Primeiro, integramos em $y,$ considerando $x$ como constante. O que restar, integramos em $x:$

$=\displaystyle\int\limits_{0}^{5}{\left.\dfrac{y^{2}}{2}\right|_{y=0}^{y=1}\,dx}+\int\limits_{0}^{5}{\left.xe^{y}\right|_{y=0}^{y=1}\,dx}\\ \\ \\ =\int\limits_{0}^{5}{\dfrac{1}{2}\,dx}+\int\limits_{0}^{5}{x(e^{1}-e^{0})\,dx}\\ \\ \\ =\dfrac{1}{2}\int\limits_{0}^{5}{dx}+(e-1)\int\limits_{0}^{5}{x\,dx}\\ \\ \\ =\dfrac{1}{2}\cdot x|_{x=0}^{x=5}+(e-1)\cdot \left.\dfrac{x^{2}}{2}\right|_{x=0}^{x=5}\\ \\ \\ =\dfrac{1}{2}\cdot 5+(e-1)\cdot \dfrac{5^{2}}{2}\\ \\ \\ =\dfrac{5}{2}+\dfrac{25\,(e-1)}{2}$