SATHEE: Indefinite Integration 1 Question 28

Indefinite Integration 1 Question 28

29. The value of the integral $\int_{e^{- 1}}^{e^{2}} \frac{\log_{e} x}{x} d x$ is

(2000, 2M)

(a) $3 / 2$

(b) $5 / 2$

(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

$\begin{aligned} = - \int_{e^{- 1}}^{1} \frac{\log_{e} x}{x} d x + \int_{1}^{e^{2}} \frac{\log_{e} x}{x} d x \\ = - \frac{1}{2} {[{(\log_{e} x)}^{2}]}_{e^{- 1}}^{1} + \frac{1}{2} {[{(\log_{e} x)}^{2}]}_{1}^{e^{2}} \\ = - \frac{1}{2} 0 - (- 1)^{2} + \frac{1}{2} (2^{2} - 0) = \frac{5}{2} \end{aligned}$

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Indefinite Integration 1 Question 28

29. The value of the integral ∫e−1e2loge⁡xxdx is

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29. The value of the integral $\int_{e^{- 1}}^{e^{2}} \frac{\log_{e} x}{x} d x$ is