Let $R$ be the region bounded by the graphs of $y = \cos \left(\frac{\pi x}{2}\right)$ and $y=x^2 -1$. The line $y=k$ splits the region $R$ into two equal parts. Find the value of $k$.
First find the area.
$$A = \int\limits_{-1}^1 \left[\cos \left(\frac{\pi x}{2}\right) – x^2 +1\right]\, \mathrm{d}x = \frac{4}{\pi} + \frac{4}{3}$$
Not exactly sure about how to find the value of $k$, whats the next step?
From the wonderful answers of both Ross Millikan and DonAntonio I have figured it out.
$$\int\limits_{-\sqrt{k+1}}^{\sqrt{k+1}} \left[k-x^2+1\right]\, \mathrm{d}x = \frac{A}{2}$$
$$\frac{4}{3}\left(k+1\right)^{3/2} = \frac{A}{2} $$
solve for $k$
$k\approx -0.015$
Best Answer
For $\,y=k\le 0\,$ evaluate the area between the line (above) and the parabola (below), and this must be half what you got in your original integral. If you get a valid value you're done, otherwise you'll have to assume $\,k>0\,$ and then you'll have to evaluate the area between the trigonometric function (above) and the line (below) as before...