The value of int_{e^{-1}}^{e^2} left| frac{lo | vedclass

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(B) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$. Since $\log_e x < 0$ for $x \in [e^{-1}, 1)$ and $\log_e x \ge 0$ for $x \in [

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$\frac{5}{2}$

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(B) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$. Since $\log_e x $I = \int_{e^{-1}}^{1} -\frac{\log_e x}{x} dx + \int_{1}^{e^2} \frac{\log_e x}{x} dx$. Let $z = \log_e x$,then $dz = \frac{1}{x} dx$. When $x = e^{-1}, z = -1$. When $x = 1, z = 0$. When $x = e^2, z = 2$. $I = \int_{-1}^{0} -z dz + \int_{0}^{2} z dz$. $I = \left[ -\frac{z^2}{2} \right]_{-1}^{0} + \left[ \frac{z^2}{2} \right]_{0}^{2}$. $I = (0 - (-1/2)) + (4/2 - 0) = \frac{1}{2} + 2 = \frac{5}{2}$.

The value of $\int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$ is

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