Question

eu quero saber a integral dupla
de 0 até pi/2 e de 0 até pi/3 de (xseny-senx) dy dx

Answer 1

$I=\displaystyle\int\limits_{0}^{\pi/2}\int\limits_{0}^{\pi/3}xsen(y)-sen(x)\,dy\,dx$

A integral de dentro, que resolvemos primeiro, está em y, logo encontraremos a antiderivada da função considerando x como constante

$I=\displaystyle\int\limits_{0}^{\pi/2}\bigg[-xcos(y)-ysen(x)\bigg]_{y=0}^{y=\pi/3}\,dx\\\\\\I=\int\limits_{0}^{\pi/2}\bigg[-xcos\left(\frac{\pi}{3}\right)-\frac{\pi}{3}sen(x)+xcos(0)+0sen(x)\bigg]\,dx\\\\\\I=\int\limits_{0}^{\pi/2}\bigg[-\dfrac{x}{2}-\dfrac{\pi}{3}sen(x)+x\bigg]\,dx\\\\\\I=\bigg[-\dfrac{~x^{2}}{4}+\dfrac{\pi}{3}cos(x)+\dfrac{x^{2}}{2}\bigg]_{0}^{\pi/2}$

$I=-\dfrac{(\frac{\pi}{2})^{2}}{4}+\dfrac{\pi}{3}cos\left(\dfrac{\pi}{2}\right)+\dfrac{(\frac{\pi}{2})^{2}}{2}+\dfrac{0^{2}}{4}-\dfrac{\pi}{3}cos(0)-\dfrac{0^{2}}{2}\\\\\\I=-\dfrac{\pi^{2}}{2^{2}\cdot4}+\dfrac{\pi}{3}\cdot0+\dfrac{\pi^{2}}{2^{2}\cdot2}+0-\dfrac{\pi}{3}-0\\\\\\I=-\dfrac{\pi^{2}}{16}+\dfrac{\pi^{2}}{8}-\dfrac{\pi}{3}\\\\\\I=\dfrac{-3\pi^{2}+6\pi^{2}-16\pi}{48}\\\\\\I=\dfrac{3\pi^{2}-16\pi}{48}\\\\\\\boxed{\boxed{I=\dfrac{\pi}{48}(3\pi-16)}}$