Definite Integration Ques 23
- The value of the integral $\int _{e^{-1}}^{e^{2}} \frac{\log _e x}{x} d x$ is
(2000, 2M)
(a) $3 / 2$
(b) $5 / 2$
(c) 3
(d) 5
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Solution:
- $\int _{e^{-1}}^{e^{2}} \frac{\log _e x}{x} d x=\int _{e^{-1}}^{1} \frac{\log _e x}{x} d x-\int _1^{e^{2}} \frac{\log _e x}{x} d x$ since, 1 is turning point for $\frac{\log _e x}{x}$ for + ve and - ve values
$$ \begin{aligned} & =-\int _{e^{-1}}^{1} \frac{\log _e x}{x} d x+\int _1^{e^{2}} \frac{\log _e x}{x} d x \\ & =-\frac{1}{2}\left[\left(\log _e x\right)^{2}\right] _{e^{-1}}^{1}+\frac{1}{2}\left[\left(\log _e x\right)^{2}\right] _1^{e^{2}} \\ & =-\frac{1}{2}{0-(-1)^{2} }+\frac{1}{2}\left(2^{2}-0\right)=\frac{5}{2} \end{aligned} $$
BETA
