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

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(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

$- 1 / 2$

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(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,$I = - 1 / 2 + 0 = - 1 / 2$.

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

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