Question

the region bounded by the x-axis and the part of the graph of $y = \cos x$ between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ is separated into two regions by the line $x = k$. if the area of the region for $-\frac{\pi}{2} \leq x \leq k$ is three times the area of the region for $k \leq x \leq \frac{\pi}{2}$, then $k = $

Explanation:

Step1: Find total area

The total area \( A_{total} \) between \( x = -\frac{\pi}{2} \) and \( x=\frac{\pi}{2} \) under \( y = \cos x \) (above x - axis) is given by the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\mathrm{d}x \). Since \( \cos x \) is even, \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\mathrm{d}x=2\int_{0}^{\frac{\pi}{2}} \cos x\mathrm{d}x \). Integrating \( \cos x \), we get \( \sin x \), so \( 2[\sin x]_{0}^{\frac{\pi}{2}} = 2(1 - 0)=2 \).

Step2: Set up area equations

Let \( A_1=\int_{-\frac{\pi}{2}}^{k}\cos x\mathrm{d}x \) and \( A_2=\int_{k}^{\frac{\pi}{2}}\cos x\mathrm{d}x \). We know \( A_1 = 3A_2 \) and \( A_1+A_2=A_{total} = 2 \). Substituting \( A_1 = 3A_2 \) into \( A_1 + A_2=2 \), we get \( 3A_2+A_2 = 2\Rightarrow4A_2 = 2\Rightarrow A_2=\frac{1}{2} \).

Step3: Solve for k

Since \( A_2=\int_{k}^{\frac{\pi}{2}}\cos x\mathrm{d}x=\frac{1}{2} \), integrating \( \cos x \) gives \( [\sin x]_{k}^{\frac{\pi}{2}}=\frac{1}{2} \). So \( \sin\frac{\pi}{2}-\sin k=\frac{1}{2} \). Since \( \sin\frac{\pi}{2} = 1 \), we have \( 1-\sin k=\frac{1}{2}\Rightarrow\sin k=\frac{1}{2} \). In the interval \( [-\frac{\pi}{2},\frac{\pi}{2}] \), \( k=\frac{\pi}{6} \) (we can check the symmetry and the integral, also since \( \cos x \) is even, and the area from \( -\frac{\pi}{2} \) to \( k \) and \( k \) to \( \frac{\pi}{2} \), solving \( \sin k=\frac{1}{2} \) in \( [-\frac{\pi}{2},\frac{\pi}{2}] \) gives \( k = \frac{\pi}{6} \) as the positive solution (and we can verify the integral for negative \( k \) would not satisfy \( A_1 = 3A_2 \) due to symmetry, but let's confirm: if \( k=-\frac{\pi}{6} \), \( A_1=\int_{-\frac{\pi}{2}}^{-\frac{\pi}{6}}\cos x\mathrm{d}x=\sin(-\frac{\pi}{6})-\sin(-\frac{\pi}{2})=-\frac{1}{2}+ 1=\frac{1}{2} \), \( A_2=\int_{-\frac{\pi}{6}}^{\frac{\pi}{2}}\cos x\mathrm{d}x=\sin\frac{\pi}{2}-\sin(-\frac{\pi}{6})=1+\frac{1}{2}=\frac{3}{2} \), which is \( A_2 = 3A_1 \), opposite of what we need. So \( k=\frac{\pi}{6} \)).

Answer:

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