91Ó°ÊÓ

In order to visualize the region we are trying to find the area of, we need to graph \(y=6 \cos x\) between \(x=-\frac{\pi}{2}\) and \(x=\pi\). The function \(y=6 \cos x\) is the cosine function with an amplitude of 6. It passes through \((-\pi/2, 0)\) and \((\pi, 0)\). The graph intersects the x-axis between the interval at \(x=0\), which divides the region into two parts: below the x-axis and above the x-axis.

To find the points where the graph intersects the x-axis, we need to find the points where \(y=0\). In other words, find the values of \(x\) where \(6\cos x = 0\) within the given interval. The cosine function is equal to zero at the points where \(x = \frac{(2n + 1)\pi}{2}\), where n is an integer. Within the interval \([-\frac{\pi}{2}, \pi]\), our points of intersection are: - \(-\frac{\pi}{2}\) which corresponds to \(n=-1\) - \(0\) which corresponds to \(n=0\) - \(\frac{\pi}{2}\) which corresponds to \(n=1\) However, since \(x=-\frac{\pi}{2}\) and \(x=\pi\) are the endpoints of the region, we only need to know the intersection at \(x=0\). This splits the integral into two intervals: \((-\frac{\pi}{2}, 0)\) and \((0, \pi)\).

The net area is the algebraic sum of the areas above and below the x-axis. We can calculate the net area by integrating the function over the two intervals we found in Step 2. The net area can be found using the following integral: $$ A_{net}=\int_{-\frac{\pi}{2}}^{0}6\cos x \,dx + \int_{0}^{\pi}6\cos x \,dx $$ Calculate the integrals: $$ A_{net} = [6\sin x]_{-\frac{\pi}{2}}^{0} + [6\sin x]_{0}^{\pi} $$ $$ A_{net} = (6\sin 0 - 6\sin(-\frac{\pi}{2})) + (6\sin \pi - 6\sin0) = 0 + 6 - 0 -0 = 6 $$ So, the net area of the region is 6.

The area of the region is the sum of the positive areas above and below the x-axis. We can find this by taking the absolute value of the integral over the two intervals we found in Step 2: $$ A = \left|\int_{-\frac{\pi}{2}}^{0}6\cos x \,dx\right| + \left|\int_{0}^{\pi}6\cos x \,dx\right| $$ Since we already calculated the integrals in Step 3, simply take their absolute values: $$ A = |(6\sin 0 - 6\sin(-\frac{\pi}{2}))| + |(6\sin \pi - 6\sin0)| = |-6| + |6|= 6 + 6 = 12 $$ Therefore, the area of the region is 12. To summarize: (i) The net area of the region is 6. (ii) The area of the region is 12.