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 $\int_{0}^{1} x^{m} (1 - x)^{n} dx = k \int_{0}^{1} x^{n} (1 - x)^{m} dx$,then the value of $k$ equals

$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$.

The value of $\int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$ is equal to

$\int_{- 1 / 2}^{1 / 2} \{ [x] + \log (\frac{1 + x}{1 - x}) \} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module