int_{- 1 / 2}^{1 / 2} { [x] + log (frac{1 + x | vedclass

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

Question

(C) Let $I = \int_{- 1 / 2}^{1 / 2} \{ [x] + \log (\frac{1 + x}{1 - x}) \} dx$. We can split the integral as $I = \int_{- 1 / 2}^{1 / 2} [x] dx + \int

Vedclass · Accepted Answer

$- 1 / 2$

Vedclass · Answer

(C) Let $I = \int_{- 1 / 2}^{1 / 2} \{ [x] + \log (\frac{1 + x}{1 - x}) \} dx$. We can split the integral as $I = \int_{- 1 / 2}^{1 / 2} [x] dx + \int_{- 1 / 2}^{1 / 2} \log (\frac{1 + x}{1 - x}) dx$. Let $f(x) = \log (\frac{1 + x}{1 - x})$. Then $f(- x) = \log (\frac{1 - x}{1 + x}) = \log (\frac{1 + x}{1 - x})^{- 1} = - \log (\frac{1 + x}{1 - x}) = - f(x)$. Since $f(x)$ is an odd function,$\int_{- 1 / 2}^{1 / 2} \log (\frac{1 + x}{1 - x}) dx = 0$. Now,consider $\int_{- 1 / 2}^{1 / 2} [x] dx$. For $x \in [- 1 / 2, 0)$,$[x] = - 1$. For $x \in [0, 1 / 2]$,$[x] = 0$. Thus,$\int_{- 1 / 2}^{1 / 2} [x] dx = \int_{- 1 / 2}^{0} (- 1) dx + \int_{0}^{1 / 2} 0 dx = [- x]_{- 1 / 2}^{0} = 0 - (1 / 2) = - 1 / 2$. Therefore,$I = - 1 / 2 + 0 = - 1 / 2$.

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

Similar Questions

$I = \int\limits_0^1 \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx$ is equal to

$\int_0^2 x^2(2-x)^5 d x=$

Let $f:R \to R$ and $g:R \to R$ be continuous functions,then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is:

If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $

The value of the integral $\int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module