91Ó°ÊÓ

To find the area between the curves, we first need to determine where they intersect. Set the equations equal: \[ x \sin x = x \]Divide by \(x\) (assuming \(x eq 0\)): \[ \sin x = 1 \]The solution in the given interval \( [0, \frac{\pi}{2}] \) is \( x = \frac{\pi}{2} \).Thus, the curves intersect at \( x = 0 \) and \( x = \frac{\pi}{2} \).

The area between the curves from \(x = 0\) to \(x = \frac{\pi}{2}\) is found by integrating the difference between the functions:\[ A = \int_{0}^{\frac{\pi}{2}} \left( x - x \sin x \right) \, dx \]

Simplify the integrand:\[ x - x \sin x = x(1 - \sin x) \]Thus, the integral simplifies to:\[ \int_{0}^{\frac{\pi}{2}} x(1 - \sin x) \, dx = \int_{0}^{\frac{\pi}{2}} x - x \sin x \, dx \]

Separate the integral into two parts:\[ \int_{0}^{\frac{\pi}{2}} x \, dx - \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \]The first integral is straightforward:\[ \int_{0}^{\frac{\pi}{2}} x \, dx = \left[ \frac{x^2}{2} \right]_0^{\frac{\pi}{2}} = \frac{\pi^2}{8} \]The second integral requires integration by parts. Let:\( u = x \) and \( dv = \sin x \, dx \).Then, \( du = dx \) and \( v = -\cos x \). Applying integration by parts, \[ \int x \sin x \, dx = -x \cos x \Big|_{0}^{\frac{\pi}{2}} + \int \cos x \, dx \]\[ = -\left( \frac{\pi}{2} \cdot 0 - 0 \cdot (-1) \right) + \sin x \Big|_{0}^{\frac{\pi}{2}} \]\[ = -\left( \frac{\pi}{2} \cdot 0 \right) + \left( 1 - 0 \right) = 1 \]Hence,\[ \int_{0}^{\frac{\pi}{2}} x \sin x \, dx = 1 \]

The area is:\[ A = \frac{\pi^2}{8} - 1 \]