Question 16

Step 1: Express the integral expression. Let $I = 2 \int_0^{\pi/2} f(x)g(x)dx - \int_0^{\pi/2} g(x)dx$. Substituting $f(x) = \sin^2 x$, we get: $I = \int_0^{\pi/2} (2 \sin^2 x - 1) g(x) dx = \int_0^{\pi/2} (-\cos 2x) g(x) dx$.

Step 2: Simplify $g(x)$. $g(x) = \sqrt{\frac{\pi x}{2} - x^2} = \sqrt{-(x^2 - \frac{\pi x}{2} + \frac{\pi^2}{16}) + \frac{\pi^2}{16}} = \sqrt{\frac{\pi^2}{16} - (x - \frac{\pi}{4})^2}$.

Step 3: Use substitution to evaluate the integral. Let $x - \frac{\pi}{4} = t \implies dx = dt$. When $x = 0, t = -\frac{\pi}{4}$. When $x = \frac{\pi}{2}, t = \frac{\pi}{4}$. Also, $\cos 2x = \cos(2(t + \frac{\pi}{4})) = \cos(2t + \frac{\pi}{2}) = -\sin 2t$.

Step 4: Substitute into the integral. $I = \int_{-\pi/4}^{\pi/4} -(-\sin 2t) \sqrt{\frac{\pi^2}{16} - t^2} dt = \int_{-\pi/4}^{\pi/4} \sin 2t \sqrt{\frac{\pi^2}{16} - t^2} dt$.

Step 5: Analyze the integrand. $\sin 2t$ is an odd function and $\sqrt{\frac{\pi^2}{16} - t^2}$ is an even function. Their product is an odd function. The integral of an odd function over a symmetric interval $[-a, a]$ is zero. Therefore, $I = 0$.