यदि int limits_0^1 frac{1}{left(5+2 x -2 x ^2 | vedclass

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(A) माना $I = \int \limits_0^1 \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$.गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करते हुए,$x$ को $1-x$ से प

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) माना $I = \int \limits_0^1 \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$.गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करते हुए,$x$ को $1-x$ से प्रतिस्थापित करने पर:$I = \int \limits_0^1 \frac{dx}{(5+2(1-x)-2(1-x)^2)(1+e^{2-4(1-x)})} = \int \limits_0^1 \frac{e^{2-4x} dx}{(5+2x-2x^2)(1+e^{2-4x})}$.दोनों समीकरणों को जोड़ने पर:$2I = \int \limits_0^1 \frac{dx}{5+2x-2x^2} = \int \limits_0^1 \frac{dx}{2(\frac{11}{4} - (x-\frac{1}{2})^2)}$.समाकलन करने पर,$I = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$.तुलना करने पर,$\alpha = \sqrt{11}$ और $\beta = \sqrt{10}$.अतः,$\alpha^4 - \beta^4 = 121 - 100 = 21$.

यदि $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ जहाँ $\alpha, \beta > 0$,तो $\alpha^4-\beta^4$ का मान ज्ञात कीजिए:

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