Cos x bounded by $x$-axis. Find k given information about k.

Question

The region bounded by the $x$-axis and the part of the graph of $y = \cos x$ between $x= -\pi/2$ and $x=\pi/2$ is separated into two regions by the line $x=k$. If the area of the region for $-\pi/2$ is less than or equal to $x$ which is less than or equal to k is three times the area of the region for $k$ is less than or equal to $x$ which is less than or equal to $\pi/2$, then $k=?$

Answer 1

Draw a picture. The area under the curve $y=\cos x$, and above the $x$-axis, from $x=-\frac{\pi}{2}$ to $x=k$ is $$\int_{-\pi/2}^k \cos x\,dx.\tag{1}$$ The area from $x=k$ to $x=\pi/2$ is $$\int_k^{\pi/2} \cos x\,dx.\tag{2}$$ The first area is $3$ times the second. Thus $$\int_{-\pi/2}^k \cos x\,dx=3\int_{k}^{\pi/2} \cos x\,dx.$$ Do the integrations. The integral (1) is $1+\sin k$, and the integral (2) is $1-\sin k$. Thus $1+\sin k=3(1-\sin k)$. It follows that $\sin k=\frac{1}{2}$ and therefore $k=\frac{\pi}{6}$.