int_{1/e}^e |log x| , dx = | vedclass

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(B) We need to evaluate the integral $I = \int_{1/e}^e |\log x| \, dx$. Since $\log x < 0$ for $x \in [1/e, 1)$ and $\log x \ge 0$ for $x \in [1, e]$,

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$2 \left( 1 - \frac{1}{e} \right)$

Vedclass · Answer

(B) We need to evaluate the integral $I = \int_{1/e}^e |\log x| \, dx$. Since $\log x $I = \int_{1/e}^1 -\log x \, dx + \int_1^e \log x \, dx$. Using the integral formula $\int \log x \, dx = x \log x - x + C$: $I = -[x \log x - x]_{1/e}^1 + [x \log x - x]_1^e$. Evaluating the first part: $-[(1 \cdot 0 - 1) - (\frac{1}{e} \log(1/e) - \frac{1}{e})] = -[-1 - (-\frac{1}{e} - \frac{1}{e})] = -[-1 + \frac{2}{e}] = 1 - \frac{2}{e}$. Evaluating the second part: $[(e \log e - e) - (1 \log 1 - 1)] = [(e - e) - (0 - 1)] = 0 - (-1) = 1$. Adding both parts: $I = (1 - \frac{2}{e}) + 1 = 2 - \frac{2}{e} = 2 \left( 1 - \frac{1}{e} \right)$.

$\int_{1/e}^e |\log x| \, dx = $

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