= Solution
The upper boundary follows $y=x$ until its intersection with $y=1/x$ at $(1,1)$, then follows the <hyperbola> to $(2,1/2)$. The lower boundary is $y=0$. Therefore the two vertical-slice descriptions are $0\leq x\leq1$, $0\leq y\leq x$, and $1\leq x\leq2$, $0\leq y\leq1/x$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1-line-hyperbola-region.png]
{title=The integration region below a line and a hyperbola}
{description=Original shaded region with all four boundary pieces shown. Vertical slices change their upper endpoint at $(1,1)$.}
Integrating with respect to $y$ first reduces the <double integral> to
$$
I=\int_0^1x^3e^{-x^2}dx+\int_1^2xe^{-x^2}dx=\left[-\frac12(x^2+1)e^{-x^2}\right]_0^1+\left[-\frac12e^{-x^2}\right]_1^2.
$$
Using the <antiderivatives> from part (a),
$$
\boxed{I=\frac{1-e^{-1}-e^{-4}}2.}
$$
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