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

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(B) આપેલ સંકલન બીટા વિધેયના સ્વરૂપમાં છે,$B(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.અહીં,$m-1 = 5/2 \impl

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ સંકલન બીટા વિધેયના સ્વરૂપમાં છે,$B(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.અહીં,$m-1 = 5/2 \implies m = 7/2$ અને $n-1 = 3/2 \implies n = 5/2$.તેથી,સંકલન $B(7/2, 5/2) = \frac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(7/2 + 5/2)} = \frac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(6)}$ છે.$\Gamma(n+1) = n\Gamma(n)$ અને $\Gamma(1/2) = \sqrt{\pi}$ ગુણધર્મનો ઉપયોગ કરતા:$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.$\Gamma(6) = 5! = 120$.આ કિંમતો મૂકતા:સંકલન $= \frac{(\frac{15\sqrt{\pi}}{8}) \cdot (\frac{3\sqrt{\pi}}{4})}{120} = \frac{45\pi}{32 \cdot 120} = \frac{45\pi}{3840} = \frac{3\pi}{256}$.

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

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