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

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(A) ધારો કે $I = \int \limits_0^1 \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$.ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,$x$ ને $1-x$ વડે બ

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) ધારો કે $I = \int \limits_0^1 \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$.ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,$x$ ને $1-x$ વડે બદલતા:$I = \int \limits_0^1 \frac{dx}{(5+2(1-x)-2(1-x)^2)(1+e^{2-4(1-x)})} = \int \limits_0^1 \frac{e^{2-4x} dx}{(5+2x-2x^2)(1+e^{2-4x})}$.બંને સમીકરણોનો સરવાળો કરતા:$2I = \int \limits_0^1 \frac{dx}{5+2x-2x^2} = \int \limits_0^1 \frac{dx}{2(\frac{11}{4} - (x-\frac{1}{2})^2)}$.સંકલન કરતા,$I = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$.સરખામણી કરતા,$\alpha = \sqrt{11}$ અને $\beta = \sqrt{10}$.તેથી,$\alpha^4 - \beta^4 = 121 - 100 = 21$.

જો $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ જ્યાં $\alpha, \beta > 0$,તો $\alpha^4-\beta^4$ ની કિંમત શોધો:

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