Question

How do i evaluate this double integral, the integrand is y^1/2*sin y dydx? x^2 < y < 1 0<x< 1

Answer 1

`= sin (1) - cos (1)`

Explanation:

Reverse the order:

`int_(y = 0)^1 \ int_(x = 0)^sqrt(y) \ y^(1/2) \ sin y \ dx \ dy`

`= int_(y = 0)^1 y\ sin y \ dy`

`= (sin y - y cos y )_0^1 = sin (1) - cos (1)`