= Solution
The <parabola> opens upward, with its minimum at $(0,-1)$ and zeros at $x=\pm1$. The quartic graph is symmetric about the vertical axis, has its flat maximum at $(0,1)$, and also has zeros at $\pm1$.
Their intersections solve $x^4+x^2-2=(x^2-1)(x^2+2)=0$, so the only real intersections are $(-1,0)$ and $(1,0)$. Between them, $1-x^4$ is the upper curve and $x^2-1$ the lower curve.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-parabola-quartic.png]
{title=The parabola, the inverted quartic, and the finite region between their intersections}
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