Question

integral dupla de 0 a π\2 de sem(x) •cos(x)y dx dy​

Answer 1

Resposta:

Lembre-se do teorema fundamental do cálculo:

$\boxed{\sf \sf\displaystyle\int^{\sf b}_{\sf a}\sf f(x)dx=F(x)\bigg|^{\sf b}_{\sf a}=F(b)-F(a)}$

Integrarei em relação a x primeiro (considere y uma constante):

$\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf (sen\,x)(cos\,x)y\,dxdy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf (sen\,x)(cos\,x)\,dx\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\displaystyle\int\sf (sen\,x)(cos\,x)dx\bigg|^{\sf\frac{\pi}{2}}_{\sf0}\bigg]dy$

Fazendo u = sen x:

$\sf u=sen\,x\implies\dfrac{du}{dx}=cos\,x\implies dx=\dfrac{1}{cos\,x}du$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\displaystyle\int\sf u(cos\,x)\bigg(\dfrac{1}{cos\,x}du\bigg)\bigg|^{\sf\frac{\pi}{2}}_{\sf0}\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\displaystyle\int\sf u\bigg(\dfrac{cos\,x}{cos\,x}du\bigg)\bigg|^{\sf\frac{\pi}{2}}_{\sf0}\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\displaystyle\int\sf u\,du\bigg|^{\sf\frac{\pi}{2}}_{\sf0}\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\dfrac{u^2}{2}\bigg|^{\sf\frac{\pi}{2}}_{\sf0}\bigg]dy$

Devolvendo sen x = u:

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\dfrac{sen^2x}{2}\bigg|^{\sf\frac{\pi}{2}}_{\sf0}\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\dfrac{sen^2\frac{\pi}{2}}{2}-\dfrac{sen^20}{2}\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\dfrac{1^2}{2}-\dfrac{0^2}{2}\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf y\bigg[\dfrac{1}{2}-0\bigg]dy$

$=\sf\displaystyle\int^{\sf\frac{\pi}{2}}_{\sf0}\sf \dfrac{y}{2}\,dy$

$=\sf\displaystyle\int\sf \dfrac{y}{2}\,dy\bigg|^{\sf\frac{\pi}{2}}_{\sf0}$

$=\sf\dfrac{y^2}{4}\bigg|^{\sf\frac{\pi}{2}}_{\sf0}$

$=\sf \dfrac{(\frac{\pi}{2})^2}{4}- \dfrac{0^2}{4}$

$=\sf \dfrac{\frac{\pi^2}{4}}{4}-0$

$\red{\sf =\dfrac{\pi^2}{16}}$