Question

Evaluate the iterated integral \int_{0}^{\pi/4} \int_{0}^{\pi/2} sin x cos 2y dx dy

Answer 1

Answer and Explanation:

Thus,

=∫π40∫π20(sin(x)(−cos(2y)))dydx[Integrate with respect to y]=∫π40(−12sin(x)sin(2y))∣∣∣π20dx=∫π40((−12sin(x)sin(2(π2)))−(−12sin(x)sin(2(0))))dx=∫π400dx

Answer 2

The integrate the iterated integral over the given rectangular region R

[0,π/4]×[0×π/2][0,π/4]×[0×π/2]

you can apply Fubini's theorem integrate ∫π/40∫π/20sin(x)cos(2y)dxdy∫0π/4∫0π/2sin(x)cos(2y)dxdy with respect to x first or y.