The value of int_{-1}^1 frac{(1+sqrt{|x|-x}) | vedclass

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

Question

(D) Let $I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx

Vedclass · Accepted Answer

$1+\frac{2 \sqrt{2}}{3}$

Vedclass · Answer

(D) Let $I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we replace $x$ with $-x$:$I = \int_{-1}^1 \frac{(1+\sqrt{|-x|-(-x)}) e^{-x}+(\sqrt{|-x|-(-x)}) e^{-(-x)}}{e^{-x}+e^{-(-x)}} dx$$I = \int_{-1}^1 \frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^x}{e^{-x}+e^x} dx$.Adding the two expressions for $I$:$2I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x + (\sqrt{|x|-x}) e^{-x} + (1+\sqrt{|x|+x}) e^{-x} + (\sqrt{|x|+x}) e^x}{e^x+e^{-x}} dx$$2I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})(e^x+e^{-x})}{e^x+e^{-x}} dx$$2I = \int_{-1}^1 (1+\sqrt{|x|-x}+\sqrt{|x|+x}) dx$.Since the integrand is an even function,$2I = 2 \int_0^1 (1+\sqrt{x-x}+\sqrt{x+x}) dx = 2 \int_0^1 (1+0+\sqrt{2x}) dx$.$I = \int_0^1 (1+\sqrt{2}x^{1/2}) dx = [x + \sqrt{2} \cdot \frac{2}{3} x^{3/2}]_0^1$$I = 1 + \frac{2\sqrt{2}}{3}$.

The value of $\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x$ is equal to

Similar Questions

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to

$\int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx =$

$\int\limits_0^\pi {{e^{{{\cos }^4}x}}} \cdot \cos^5(2n + 1)x \,dx, (n \in I)$ is equal to

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$

If $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 (3 - f(x)) dx = 7$,then $\int_{-1}^2 f(x) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module