First slide

Evaluation of definite integrals

Question

Given $I_{m} = \int_{1}^{e} (\log x)^{m} d x$ , If $\frac{I_{m}}{K} + \frac{I_{m - 2}}{L} = e$ then values of K and L are

Moderate

A

$1 - m, \frac{1}{m}$
B

$\frac{1}{1 - m}, m$
C

$\frac{1}{1 - m}, \frac{m (m - 2)}{m - 1}$
D

$\frac{m}{m - 1}, m - 2$

Solution

$I_{m} = \int_{1}^{e} (\log x)^{m} d x = {x (\log x)^{m}|}_{1}^{e} - m \int_{1}^{e} (\log x)^{m - 1} d x$
$\begin{array}{l} = e - m [{x (\log x)^{m - 1}|}_{1}^{e} - (m - 1) \int_{1}^{e} (\log x)^{m - 2} d x] \\ = e - m e + m (m - 1) I_{m - 2} = (1 - m) e + m (m - 1) I_{m - 2} \end{array}$
So $\frac{I_{m}}{1 - m} + m I_{m - 2} = e$ . Thus $K = 1 - m$ and $L = \frac{1}{m}$ .

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