$Let I= \int\limits_0^5\int\limits_{2-x}^{2+x} dxdy$

Units of x are $x=0 \space to \space x=5$

Limits of y are $y=2-x \space \&\space y=2+x$ which is equation of line

1st equation $y=2-x$

When $x=0 , y=2 $

$X=2 ,y=0 $

$\therefore$ Line passes from $(0,2) \space \&\space (2,0)$

2nd equation $y=2+x$

When $x=0 ,y=2 $

$x=2 ,y=4$

$\therefore$ Line passes from $(0,2) \space \&\space (2,4)$

$\therefore$ Intersection of $x=5 \space \&\space y=2+x$

$\therefore y=7$ & $x=5$

$\therefore (5,7)$

Intersection of $x=5$ and $y=2-x$

$\therefore y=-3 ,x=5 $

$\therefore (5,-3)$ $\therefore$ Now changing the order the region is divided into two parts as upper limits changes at $y=2$

Region (1)

1) Outer limit $y=-3$ to $y=2$

2) Inner limit is x

a) Lower limit is equation of line $y=2-x$

$\therefore x=2-y$

b) Upper limit is $x=5$

Region (2)

1) Outer limit is $y=2$ to 7

2) Inner limit x

a) lower limit is equation of line

$ y=2+x \therefore x=y-2$

b) Upper limit $x=5 $

$\therefore I=\int\limits_{-3}^2\int\limits^5_{2-y}dxdy+ \int\limits_{2}^7\int\limits^5_{y-2}dxdy\\ =\int\limits_{-3}^2[x]^5_{2-y}dy+\int\limits_{2}^7[x]^5_{y-2}dy\\ I= \int\limits_{-3}^2[5-(2-y)]dy+ \int\limits_{2}^7[5-(y-2)]dy\\ =\int\limits_{-3}^2(3+y)dy+\int\limits_{2}^7(7-y)dy\\ =\Bigg[3y+\dfrac {y^2}2\Bigg]^2_{-3}+\Bigg[7y-\dfrac {y^2}2\Bigg]^7_2\\ =\Bigg(6+\dfrac 42\Bigg)-\Bigg(-9+\dfrac 92\Bigg)+\Bigg(49-\dfrac {49}2\Bigg)-\dfrac (14-\dfrac 72\Bigg)\\ =8+\dfrac 92+\dfrac {49}2-12\\ =25$