int_0^2 x^{frac{5}{2}} sqrt{2-x} , dx = | vedclass

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(C) ધારો કે $I = \int_0^2 x^{5/2} \sqrt{2-x} \, dx$. $x = 2 \sin^2 	heta$ આદેશ લેતા, $dx = 4 \sin 	heta \cos 	heta \, d	heta$ મળે. જ્યારે $x=0,

Vedclass · Accepted Answer

$\frac{5 \pi}{8}$

Vedclass · Answer

(C) ધારો કે $I = \int_0^2 x^{5/2} \sqrt{2-x} \, dx$. $x = 2 \sin^2 	heta$ આદેશ લેતા, $dx = 4 \sin 	heta \cos 	heta \, d	heta$ મળે. જ્યારે $x=0, 	heta=0$ અને જ્યારે $x=2, 	heta=\pi/2$ થાય. $I = \int_0^{\pi/2} (2 \sin^2 	heta)^{5/2} \sqrt{2 - 2 \sin^2 	heta} \cdot (4 \sin 	heta \cos 	heta) \, d	heta$ $I = \int_0^{\pi/2} (2^{5/2} \sin^5 	heta) (\sqrt{2} \cos 	heta) (4 \sin 	heta \cos 	heta) \, d	heta$ $I = \int_0^{\pi/2} (2^{5/2} \cdot 2^{1/2} \cdot 4) \sin^6 	heta \cos^2 	heta \, d	heta$ $I = 32 \int_0^{\pi/2} \sin^6 	heta \cos^2 	heta \, d	heta$ વોલિસના સૂત્રનો ઉપયોગ કરતા: $I = 32 \cdot \frac{(5 \cdot 3 \cdot 1) \cdot (1)}{(8 \cdot 6 \cdot 4 \cdot 2)} \cdot \frac{\pi}{2}$ $I = 32 \cdot \frac{15}{384} \cdot \frac{\pi}{2} = 32 \cdot \frac{5}{128} \cdot \frac{\pi}{2} = \frac{5 \pi}{8}$.

$\int_0^2 x^{\frac{5}{2}} \sqrt{2-x} \, dx =$

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