Question:
If $ f(x)={{\log }_{e}}\,(1+x)-{{\log }_{e}}(1-x), $ then the value of $ \int_{-1/2}^{1/2}{f(x)\,\,\,dx} $ equals to
If $ f(x)={{\log }_{e}}\,(1+x)-{{\log }_{e}}(1-x), $ then the value of $ \int_{-1/2}^{1/2}{f(x)\,\,\,dx} $ equals to
Updated On: Jun 23, 2024
- $ 0 $
- $ 1 $
- $ \frac{1}{2} $
- $ -\frac{1}{2} $
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The Correct Option is A
Solution and Explanation
Given, $ f(x)={{\log }_{e}}(1+x)-{{\log }_{e}}(1-x) $
$ \therefore $ $ f(-x)={{\log }_{e}}(1-x)-{{\log }_{e}}(1+x) $
$ =-[{{\log }_{e}}(1+x)-{{\log }_{e}}(1-x)] $
$ =-f(x) $
$ \therefore $ $ f(x) $ is an odd function.
$ \therefore $ $ \int_{-1/2}^{1/2}{f(x)\,dx=0} $
$ \therefore $ $ f(-x)={{\log }_{e}}(1-x)-{{\log }_{e}}(1+x) $
$ =-[{{\log }_{e}}(1+x)-{{\log }_{e}}(1-x)] $
$ =-f(x) $
$ \therefore $ $ f(x) $ is an odd function.
$ \therefore $ $ \int_{-1/2}^{1/2}{f(x)\,dx=0} $
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
