If f(x) = frac{e^x}{1 + e^x},I_1 = int_{f(-a) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = \frac{e^x}{1 + e^x}$. Note that $f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1}$.Thus,$f(a) + f(-a) = \frac{e^a}{1 + e^a} + \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given $f(x) = \frac{e^x}{1 + e^x}$. Note that $f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1}$.Thus,$f(a) + f(-a) = \frac{e^a}{1 + e^a} + \frac{1}{1 + e^a} = 1$.Let $I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$. Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A + B - x) dx$,where $A + B = 1$:$I_1 = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)(1 - (1 - x))\} dx = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)x\} dx$.$I_1 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx - \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$.$I_1 = I_2 - I_1$.$2I_1 = I_2$.Therefore,$\frac{I_2}{I_1} = 2$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

If $f(x) = \frac{e^x}{1 + e^x}$,$I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$,and $I_2 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx$,then the value of $\frac{I_2}{I_1}$ is

Similar Questions

$\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) d x=$

Evaluate the integral: $\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x$

If $\int_{0}^{1} \tan ^{-1} x \, dx = p$,then the value of $\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$ is

$\int_0^{2 \pi} \sin ^3 x \cos ^2 x \, dx = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module