ધારો કે f: [0, frac{pi}{2}] rightarrow [0, 1] | vedclass

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

Vedclass · Accepted Answer

$0, 0.25$

Vedclass · Answer

(B) $(1)$ ધારો કે $I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{\frac{\pi x}{2} - x^2} dx - \int_0^{\frac{\pi}{2}} \sqrt{\frac{\pi x}{2} - x^2} dx$.ધારો કે $I_1 = \int_0^{\frac{\pi}{2}} \sin^2 x \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.ગુણધર્મ $\int_0^a h(x) dx = \int_0^a h(a-x) dx$ નો ઉપયોગ કરતા:$I_1 = \int_0^{\frac{\pi}{2}} \cos^2 x \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx$.બંને સમીકરણોનો સરવાળો કરતા:$2I_1 = \int_0^{\frac{\pi}{2}} (\sin^2 x + \cos^2 x) \sqrt{(\frac{\pi}{4})^2 - (x - \frac{\pi}{4})^2} dx = \int_0^{\frac{\pi}{2}} g(x) dx$.આમ,$I = 2I_1 - \int_0^{\frac{\pi}{2}} g(x) dx = 0$.$(2)$ ઉપર મુજબ,$I_1 = \frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) dx = \frac{\pi^3}{64}$.તેથી,$\frac{16}{\pi^3} I_1 = \frac{16}{\pi^3} \cdot \frac{\pi^3}{64} = 0.25$.

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