please solve this double integral?

The idea is solving the indefinite integral first, then plug in the bound at the end of your calculation.

∫�� e^(-0.5(x+y)) dy dx

= ∫∫ e^(-0.5x -0.5y) dy dx

= ∫∫ e^(-0.5x).e^(-0.5y) dy dx

= ∫ e^(-0.5x) (∫ e^(-0.5y) dy) dx

∫ e^(-0.5y) dy

Suppose -0.5 y= p, so: -0.5 dy= dp, then:

1/(-0.5)∫ -0.5*e^(-0.5y) dy

= 1/(-0.5) ∫ e^p dp

= -2 (e^p) +C

= -2e^(-0.5y) +C

From y=0 to x, we got:

= -2e^(-0.5x) + 2e^(0), then:

∫ e^(-0.5x)*(-2e^(-0.5x)+2) dx

= ∫ 2e^(-0.5x) -2e^(-x) dx

The integral ∫ 2e^(-0.5 x) dx is:

-2*(1/0.5) ∫ 0.5*e^(-0.5x) dx

= -4(e^(-0.5x)) +C

and the integral ∫-2e^(-x) dx is:

= 2∫ -e^(-x) dx

= 2*e^(-x) +C

Substituting the value and we got:

= ∫ 2e^(-0.5x) -2e^(-x) dx

= -4(e^(-0.5x)) +2e^(-x) +C

For x=0 to infinity we got:

= -4(e^inf) -2e^inf -(-4e^(0) +2e^(0))

= (0)-(-4+2)

= -(-2)

= 2