= Solution
The <region of integration> is $0\leq y\leq1$, $y\leq x\leq2-y$. It is the closed <triangle> with vertices $(0,0)$, $(2,0)$, and $(1,1)$, bounded by $y=0$, $x=y$, and $x=2-y$. Each horizontal slice runs from the rising left edge to the falling right edge.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2-q8-triangular-region.png]
{title=Triangular integration region and a horizontal integration slice}
{description=Original diagram of the region $0\leq y\leq1$, $y\leq x\leq2-y$, with its three vertices, boundary equations, and an example slice at fixed $y$.}
Integrate with respect to $x$ first:
$$
\begin{aligned}
I&=\int_0^1\left[\frac23x^3+yx\right]_{x=y}^{x=2-y}\,dy\\
&=\int_0^1\left[\frac{16}3-6y+2y^2-\frac43y^3\right]dy\\
&=\frac{16}3-3+\frac23-\frac13=\boxed{\frac83}.
\end{aligned}
$$
The <triangle> has area one, but the nonconstant integrand makes the <double integral> a weighted area.
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