int_{1/4}^{1/2} frac{dx}{sqrt{x - x^2}} = | vedclass

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(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.We can rewrite the expression inside the square root by completing the square:$x - x^2 = -(x^

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.We can rewrite the expression inside the square root by completing the square:$x - x^2 = -(x^2 - x) = -((x - 1/2)^2 - 1/4) = (1/2)^2 - (x - 1/2)^2$.Thus,$I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{(1/2)^2 - (x - 1/2)^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get:$I = [\sin^{-1}(\frac{x - 1/2}{1/2})]_{1/4}^{1/2} = [\sin^{-1}(2x - 1)]_{1/4}^{1/2}$.Evaluating at the limits:$I = \sin^{-1}(2(1/2) - 1) - \sin^{-1}(2(1/4) - 1) = \sin^{-1}(0) - \sin^{-1}(-1/2)$.$I = 0 - (-\pi/6) = \pi/6$.

$\int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}} = $

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