First slide

Evaluation of definite integrals

Question

The value of $\int_{e^{- 1}}^{e^{2}} |\frac{\log x}{x}| d x$ is

Easy

A

3/2
B

5/2
C

3
D

5

Solution

$\int_{e^{- 1}}^{e^{2}} |\frac{\log x}{x}| d x = - \int_{e^{- 1}}^{1} \frac{\log x}{x} d x + \int_{1}^{e^{2}} \frac{\log x}{x} d x$
$(since \log x < 0 for x \in [e^{- 1}, 1] and \log x > 0 for x \in (1, e^{2}))$
$= - \int_{- 1}^{0} t d t + \int_{0}^{2} t d t = - {\frac{t^{2}}{2}|}_{- 1}^{0} + {\frac{t^{2}}{2}|}_{0}^{2} = \frac{1}{2} + 2 = \frac{5}{2}$

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