$X=1, y=1$

$X+y=1$

Let $x=0$ then $y=1 $

$x=1$ then $y=0 $

$\therefore x+y=1$ is a line passing from $(0,1) \space \&\space (1,0)$

1) outer limit is $x=0$ to $x=1 $

2) Inner limit y

Upper limit is $y=1 $

Lower limit is equation of line

$x+y=1 \therefore y=1-x \\ \therefore I=\int\limits_{x=0}^{x=1} \int\limits_{y=1-x}^{y=1}e^{2x-3y}dxdy \\ =\int\limits_0^1 e^{2x} \int\limits_{y=1-x}^{y=1} e^{-3y}dydx \\ =\int\limits_0^1 e^{2x} \Bigg[\dfrac {e^{-3y}}{-3}\Bigg]_{1-x}^1 dx \\ I=\dfrac {-1}3 \int\limits_0^1 [e^{2x} [e^{-3} -e^{-3(-x)}]] dx\\ =\dfrac{-1}3\int\limits_0^1\Bigg[\dfrac {e^{2x}}{e^3}-e^{-3}.e^{5x}\Bigg]dx\\ =\dfrac {-1}{3e^3}\int\limits_0^1[e^{2x}-e^{5x}]dx\\ =\dfrac {-1}{3e^3}\Bigg[\dfrac {e^{2x}}{2}-\dfrac {e^{5x}}5\Bigg]^1_0\\ =\dfrac {-1}{3e^3}\Bigg\{\Bigg[\dfrac {e^2}2-\dfrac {e^5}5\Bigg]-\Bigg[\dfrac 12-\dfrac 15\Bigg]\Bigg\}\\ =\dfrac {-1}{3e^3}\Bigg[\dfrac {e^2}2-\dfrac {e^5}5-\dfrac 3{10}\Bigg]\\ =\dfrac {-1}{30}[5e^{-1}-2e^2-3e^{-3}] $