[tex]I=\displaystyle\iint_D\mathrm dA=\int_{x=0}^{x=4}\int_{y=0}^{y=4-x}\mathrm dy\,\mathrm dx[/tex]
[tex]I=\displaystyle\int_{x=0}^{x=4}(4-x)\,\mathrm dx[/tex]
[tex]I=8[/tex]
To verify this, we can simply find the area of the triangle using the well-known formula [tex]\dfrac12bh[/tex], where [tex]b[/tex] and [tex]h[/tex] are the base and height, respectively of the triangular region [tex]D[/tex]. We have [tex]b=h=4[/tex], so the area is [tex]\dfrac{4^2}2=8[/tex], as expected.
