The value of the integral int_{-1}^{1} log le | vedclass

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

Question

(B) Let $I = \int_{-1}^{1} f(x) \, dx$,where $f(x) = \log \left(x+\sqrt{x^{2}+1}\right)$.Check if $f(x)$ is an odd function by evaluating $f(-x)$:$f(-

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $I = \int_{-1}^{1} f(x) \, dx$,where $f(x) = \log \left(x+\sqrt{x^{2}+1}\right)$.Check if $f(x)$ is an odd function by evaluating $f(-x)$:$f(-x) = \log \left(-x+\sqrt{(-x)^{2}+1}\right) = \log \left(\sqrt{x^{2}+1}-x\right)$.Multiply and divide by $(\sqrt{x^{2}+1}+x)$:$f(-x) = \log \left(\frac{(\sqrt{x^{2}+1}-x)(\sqrt{x^{2}+1}+x)}{\sqrt{x^{2}+1}+x}\right) = \log \left(\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}+x}\right) = \log \left(\frac{1}{\sqrt{x^{2}+1}+x}\right)$.Using the property $\log(1/a) = -\log(a)$:$f(-x) = -\log \left(x+\sqrt{x^{2}+1}\right) = -f(x)$.Since $f(-x) = -f(x)$,the function 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,$\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) \, dx = 0$.

The value of the integral $\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) \, dx$ is:

Similar Questions

If $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,then $k =$

The value of the integral $\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x$ is equal to

$\int_0^{2 \pi} \sin ^6 x \cos ^5 x \, dx$ is equal to

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

The value of $\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module