Question 17

Step 1: Use the result from the previous question. From the solution of Q.16, we know that $2 \int_0^{\pi/2} f(x)g(x)dx - \int_0^{\pi/2} g(x)dx = 0$. This implies $2 \int_0^{\pi/2} f(x)g(x)dx = \int_0^{\pi/2} g(x)dx$, so $\int_0^{\pi/2} f(x)g(x)dx = \frac{1}{2} \int_0^{\pi/2} g(x)dx$.

Step 2: Calculate $\int_0^{\pi/2} g(x)dx$. $\int_0^{\pi/2} g(x)dx = \int_0^{\pi/2} \sqrt{\frac{\pi^2}{16} - (x - \frac{\pi}{4})^2} dx$. Let $x - \frac{\pi}{4} = t \implies dx = dt$. Integral becomes $\int_{-\pi/4}^{\pi/4} \sqrt{(\frac{\pi}{4})^2 - t^2} dt$. This represents the area of a semi-circle with radius $R = \frac{\pi}{4}$. Area $= \frac{1}{2} \pi R^2 = \frac{1}{2} \pi (\frac{\pi}{4})^2 = \frac{\pi^3}{32}$.

Step 3: Calculate $\int_0^{\pi/2} f(x)g(x)dx$. $\int_0^{\pi/2} f(x)g(x)dx = \frac{1}{2} (\frac{\pi^3}{32}) = \frac{\pi^3}{64}$.

Step 4: Find the final value. Value $= \frac{16}{\pi^3} \cdot \frac{\pi^3}{64} = \frac{16}{64} = 0.25$.