Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :
y = -1 - x ==> x = -1 - y
y = x - 1 ==> x = 1 + y
So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.
This means the area is given by the integral,
[tex]\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy[/tex]
The integral with respect to x is trivial:
[tex]\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy[/tex]
For the remaining integral, integrate term-by-term to get
[tex]\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}[/tex]
Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.
