int_{e^{-1}}^{e^2} left| frac{log x}{x} right | vedclass

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

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(D) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx$. Since $\log x $I = \int_{e^{-1}}^{1} \left( -\frac{\log x}{x} \right) dx + \int_{1}^{e^2} \left( \frac{\log x}{x} \right) dx$. Substitute $t = \log x$,so $dt = \frac{dx}{x}$. When $x = e^{-1}, t = -1$. When $x = 1, t = 0$. When $x = e^2, t = 2$. $I = -\int_{-1}^{0} t dt + \int_{0}^{2} t dt$. $I = -\left[ \frac{t^2}{2} \right]_{-1}^{0} + \left[ \frac{t^2}{2} \right]_{0}^{2}$. $I = -\left( 0 - \frac{(-1)^2}{2} \right) + \left( \frac{2^2}{2} - 0 \right)$. $I = -\left( -\frac{1}{2} \right) + \frac{4}{2} = \frac{1}{2} + 2 = \frac{5}{2}$. Thus,the correct option is $D$.

$\int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx =$

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