int_{-1}^1 frac{sqrt{1+x+x^2}-sqrt{1-x+x^2}}{ | vedclass

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

Question

(C) Let $I = \int_{-1}^1 f(x) dx$,where $f(x) = \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$.We check if the function is even

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $I = \int_{-1}^1 f(x) dx$,where $f(x) = \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \frac{\sqrt{1+(-x)+(-x)^2}-\sqrt{1-(-x)+(-x)^2}}{\sqrt{1+(-x)+(-x)^2}+\sqrt{1-(-x)+(-x)^2}}$$f(-x) = \frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}}$$f(-x) = -\left( \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} \right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) dx = 0$.Therefore,$\int_{-1}^1 f(x) dx = 0$.

$\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} dx$ is equal to

Similar Questions

If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then

$\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt =$

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx =$

$\int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x=$

$\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module