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 $x \in (1, e^{2})$

= $\int_{- 1}^{0} t d t + - \int_{1}^{2} t d t = - \frac{t^{2}}{2} {|_{- 1}^{0} + \frac{t^{2}}{2} |}_{0}^{2} = \frac{1}{2} + 2 = \frac{5}{2}$

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