int_0^1 x^{5/2} (1-x)^{3/2} , dx = | vedclass

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(B) The given integral is of the form of the Beta function,$B(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.Her

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$\frac{3\pi}{256}$

Vedclass · Answer

(B) The given integral is of the form of the Beta function,$B(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.Here,$m-1 = 5/2 \implies m = 7/2$ and $n-1 = 3/2 \implies n = 5/2$.So,the integral is $B(7/2, 5/2) = \frac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(7/2 + 5/2)} = \frac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(6)}$.Using the property $\Gamma(n+1) = n\Gamma(n)$ and $\Gamma(1/2) = \sqrt{\pi}$:$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.$\Gamma(6) = 5! = 120$.Substituting these values:Integral $= \frac{(\frac{15\sqrt{\pi}}{8}) \cdot (\frac{3\sqrt{\pi}}{4})}{120} = \frac{45\pi}{32 \cdot 120} = \frac{45\pi}{3840} = \frac{3\pi}{256}$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$\int_0^1 x^{5/2} (1-x)^{3/2} \, dx =$

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