Question

the region bounded by the x-axis and the part of the graph of y = cosx between x = -π/2 and x = π/2 is separated into two regions by the line x = k. if the area of the region for -π/2 ≤ x ≤ k is three times the area of the region for k ≤ x ≤ π/2, then k =

Explanation:

Step1: Calculate the integrals

The area \(A\) under the curve \(y = \cos x\) from \(a\) to \(b\) is given by \(A=\int_{a}^{b}\cos xdx=\sin x\big|_{a}^{b}=\sin b-\sin a\).
The area of the region for \(-\frac{\pi}{2}\leq x\leq k\) is \(A_1=\int_{-\frac{\pi}{2}}^{k}\cos xdx=\sin k-\sin(-\frac{\pi}{2})=\sin k + 1\).
The area of the region for \(k\leq x\leq\frac{\pi}{2}\) is \(A_2=\int_{k}^{\frac{\pi}{2}}\cos xdx=\sin\frac{\pi}{2}-\sin k=1 - \sin k\).

Step2: Set up the equation

Since \(A_1 = 3A_2\), we have \(\sin k+1=3(1 - \sin k)\).

Step3: Solve the equation

Expand the right - hand side: \(\sin k + 1=3-3\sin k\).
Add \(3\sin k\) to both sides: \(4\sin k+1 = 3\).
Subtract \(1\) from both sides: \(4\sin k=2\).
Divide both sides by \(4\): \(\sin k=\frac{1}{2}\).
Since \(-\frac{\pi}{2}\leq k\leq\frac{\pi}{2}\), then \(k=\frac{\pi}{6}\).

Answer:

\(\frac{\pi}{6}\)