int_{1/5}^{1/2} frac{sqrt{x-x^2}}{x^3} dx = | vedclass

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(B) Let $I = \int_{1/5}^{1/2} \frac{\sqrt{x-x^2}}{x^3} dx$. We can rewrite the integrand as $\frac{\sqrt{x^2(\frac{1}{x}-1)}}{x^3} = \frac{x\sqrt{\fra

Vedclass · Accepted Answer

$\frac{14}{3}$

Vedclass · Answer

(B) Let $I = \int_{1/5}^{1/2} \frac{\sqrt{x-x^2}}{x^3} dx$. We can rewrite the integrand as $\frac{\sqrt{x^2(\frac{1}{x}-1)}}{x^3} = \frac{x\sqrt{\frac{1}{x}-1}}{x^3} = \frac{\sqrt{\frac{1}{x}-1}}{x^2}$. Let $u = \frac{1}{x} - 1$. Then $du = -\frac{1}{x^2} dx$,which implies $-\frac{1}{x^2} dx = du$. When $x = 1/5$,$u = 5 - 1 = 4$. When $x = 1/2$,$u = 2 - 1 = 1$. Substituting these into the integral: $I = \int_{4}^{1} \sqrt{u} (-du) = \int_{1}^{4} u^{1/2} du$. Evaluating the integral: $I = [\frac{u^{3/2}}{3/2}]_{1}^{4} = \frac{2}{3} [u^{3/2}]_{1}^{4}$. $I = \frac{2}{3} (4^{3/2} - 1^{3/2}) = \frac{2}{3} (8 - 1) = \frac{2}{3} 	imes 7 = \frac{14}{3}$.

$\int_{1/5}^{1/2} \frac{\sqrt{x-x^2}}{x^3} dx =$

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