Let f: [0, frac{pi}{2}] rightarrow [0, 1] be | vedclass

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(B) $(1)$ Let $I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{\frac{\pi x}{2} - x^2} dx - \int_0^{\frac{\pi}{2}} \sqrt{\frac{\pi x}{2} - x^2} dx$.Let $I_

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$0, 0.25$

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(B) $(1)$ Let $I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{\frac{\pi x}{2} - x^2} dx - \int_0^{\frac{\pi}{2}} \sqrt{\frac{\pi x}{2} - x^2} dx$.Let $I_1 = \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.Using the property $\int_0^a h(x) dx = \int_0^a h(a-x) dx$,we get:$I_1 = \int_0^{\frac{\pi}{2}} \cos^2 x \sqrt{(\frac{\pi}{4})^2 - (\frac{\pi}{2} - x - \frac{\pi}{4})^2} dx = \int_0^{\frac{\pi}{2}} \cos^2 x \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.Adding the two expressions for $I_1$:$2I_1 = \int_0^{\frac{\pi}{2}} (\sin^2 x + \cos^2 x) \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx = \int_0^{\frac{\pi}{2}} g(x) dx$.Thus,$I = 2I_1 - \int_0^{\frac{\pi}{2}} g(x) dx = \int_0^{\frac{\pi}{2}} g(x) dx - \int_0^{\frac{\pi}{2}} g(x) dx = 0$.$(2)$ From above,$I_1 = \frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.Using $\int \sqrt{a^2 - u^2} du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}(\frac{u}{a}) + C$:$I_1 = \frac{1}{2} [\frac{x - \frac{\pi}{4}}{2} \sqrt{\frac{\pi x}{2} - x^2} + \frac{\pi^2}{32} \sin^{-1}(\frac{x - \frac{\pi}{4}}{\pi/4})]_0^{\frac{\pi}{2}} = \frac{1}{2} [(\frac{\pi^2}{32} \cdot \frac{\pi}{2}) - (\frac{\pi^2}{32} \cdot -\frac{\pi}{2})] = \frac{1}{2} [\frac{\pi^3}{64} + \frac{\pi^3}{64}] = \frac{\pi^3}{64}$.Therefore,$\frac{16}{\pi^3} I_1 = \frac{16}{\pi^3} \cdot \frac{\pi^3}{64} = \frac{1}{4} = 0.25$.

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