int_0^1 frac{x^7}{sqrt{1 - x^4}} dx का मान ज् | vedclass

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Question

(B) माना $I = \int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{x^6 \cdot x}{\sqrt{1 - x^4}} dx$.$u = x^4$ प्रतिस्थापित करने पर,$du = 4x^3 dx$।

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) माना $I = \int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{x^6 \cdot x}{\sqrt{1 - x^4}} dx$.$u = x^4$ प्रतिस्थापित करने पर,$du = 4x^3 dx$। अतः $x^4 = u$ और $x^2 = \sqrt{u}$।$I = \int_0^1 \frac{x^4 \cdot x^3}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{u}{\sqrt{1 - u}} \cdot \frac{1}{4} du$.$1 - u = v^2$ लेने पर,$du = -2v dv$। जब $u=0, v=1$; जब $u=1, v=0$।$I = \frac{1}{4} \int_1^0 \frac{1 - v^2}{v} (-2v) dv = \frac{1}{2} \int_0^1 (1 - v^2) dv$.$I = \frac{1}{2} [v - \frac{v^3}{3}]_0^1 = \frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.

$\int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx$ का मान ज्ञात कीजिए।

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