int_{-4}^{4} log left(frac{8-x}{8+x}right) d | vedclass

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

Question

(D) Let $f(x) = \log \left(\frac{8-x}{8+x}\right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left(\frac{8-(-x)}{8+(

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $f(x) = \log \left(\frac{8-x}{8+x}\right)$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \log \left(\frac{8-(-x)}{8+(-x)}\right) = \log \left(\frac{8+x}{8-x}\right)$.Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:$f(-x) = -\log \left(\frac{8-x}{8+x}\right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) d x = 0$.Therefore,$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x = 0$.

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

Similar Questions

By using the properties of definite integrals,evaluate the integral $\int_{0}^{2} x \sqrt{2-x} \, dx$.

$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=$

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ where $\alpha, \beta > 0$,then $\alpha^4-\beta^4$ is equal to:

The value of the integral $\int \limits_{1 / 2}^2 \frac{\tan ^{-1} x}{x} d x$ is equal to

The value of $\int_0^\pi \sin^{50} x \cos^{49} x \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module