Question

Determine o valor da integral dupla retangular de f(x,y) = y.cos(xy) sobre o retângulo
[0,pi] x [0,1].

a) 1/2
b) n/2
c) 2/n
d) 2
e) n

Answer 1

Olá!

$\displaystyle \mathsf{ \iint_R y \, cos(yx) } \, \, , \, \, \mathsf{onde \, \, \, \, R = [0, \pi] \times [0,1] }$

Temos:

$\displaystyle \mathsf{ \int_{0}^{1} \int_{0}^{\pi} y \, cos(yx) \, \, dx \, \, dy }$

Vamos resolver a integral de dentro, considerando y como sendo uma constante, ou seja, um número qualquer.

$\displaystyle \mathsf{\int_{0}^{\pi} y \, cos(yx) \, \, dx } \\ \\ \\ \mathsf{ y \cdot \int_{0}^{\pi} cos(yx) \, \, dx} \\ \\ \\ \mathsf{ y \cdot \bigg( \frac{1}{y} sen(yx) \bigg) } \\ \\ \\ \mathsf{ sen(yx) \, \bigg|_{\, \displaystyle x=0}^{\, \displaystyle x=\pi} } \\ \\ \\ \mathsf{sen(\pi y)}$

Daí vamos continuar com a integral de fora:

$\displaystyle \mathsf{\int_{0}^{1} sen(\pi y) \, \, dy} \\ \\ \\ \mathsf{-\frac{1}{\pi} cos(\pi y) \, \bigg|_{ \, \displaystyle y=0}^{\, \displaystyle y=1} } \\ \\ \\ \mathsf{\bigg( -\frac{1}{\pi}cos(\pi) \bigg) - \bigg( -\frac{1}{\pi}cos(0) \bigg)} \\ \\ \\ \mathsf{-\frac{1}{\pi} \cdot (-1) + \frac{1}{\pi} \cdot 1} \\ \\ \\ \mathsf{\frac{1}{\pi}+\frac{1}{\pi}} \\ \\ \\ \mathsf{\boxed{\boxed{ V = \frac{2}{\pi} }}}$