int_{alpha}^{beta} sqrt{frac{x-alpha}{beta-x} | vedclass

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Question

(C) ધારો કે $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.$x = \alpha \cos^2 t + \beta \sin^2 t$ આદેશ લેતા.તેથી $dx = 2(\beta - \alph

Vedclass · Accepted Answer

$\frac{\pi}{2}(\beta - \alpha)$

Vedclass · Answer

(C) ધારો કે $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.$x = \alpha \cos^2 t + \beta \sin^2 t$ આદેશ લેતા.તેથી $dx = 2(\beta - \alpha) \sin t \cos t dt$ મળે.જ્યારે $x = \alpha$,ત્યારે $t = 0$. જ્યારે $x = \beta$,ત્યારે $t = \frac{\pi}{2}$.$x - \alpha = (\beta - \alpha) \sin^2 t$ અને $\beta - x = (\beta - \alpha) \cos^2 t$ થાય.તેથી,$\sqrt{\frac{x-\alpha}{\beta-x}} = \sqrt{\frac{(\beta-\alpha)\sin^2 t}{(\beta-\alpha)\cos^2 t}} = 	an t$.$I = \int_{0}^{\pi/2} 	an t \cdot 2(\beta - \alpha) \sin t \cos t dt = 2(\beta - \alpha) \int_{0}^{\pi/2} \sin^2 t dt$.$\sin^2 t = \frac{1 - \cos 2t}{2}$ નિત્યસમનો ઉપયોગ કરતા:$I = 2(\beta - \alpha) \int_{0}^{\pi/2} \frac{1 - \cos 2t}{2} dt = (\beta - \alpha) [t - \frac{\sin 2t}{2}]_{0}^{\pi/2}$.$I = (\beta - \alpha) [\frac{\pi}{2} - 0] = \frac{\pi}{2}(\beta - \alpha)$.

$\int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$ ની કિંમત શોધો.

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