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

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Question

(A) Let $I = \int_{-1/2}^{1/2} [x] dx + \int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx$. Consider the function $f(x) = \log \left( \frac{1+x

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int_{-1/2}^{1/2} [x] dx + \int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx$. Consider the function $f(x) = \log \left( \frac{1+x}{1-x} \right)$. Since $f(-x) = \log \left( \frac{1-x}{1+x} \right) = -\log \left( \frac{1+x}{1-x} \right) = -f(x)$,$f(x)$ is an odd function. Therefore,$\int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx = 0$. Now,$I = \int_{-1/2}^{1/2} [x] dx$. In the interval $[-1/2, 0)$,$[x] = -1$. In the interval $[0, 1/2]$,$[x] = 0$. Thus,$I = \int_{-1/2}^{0} (-1) dx + \int_{0}^{1/2} (0) dx = -[x]_{-1/2}^{0} = -(0 - (-1/2)) = -1/2$.

The integral $\int_{-1/2}^{1/2} \left( [x] + \log \left( \frac{1+x}{1-x} \right) \right) dx$ is equal to (where $[.]$ is the greatest integer function):

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