First slide

Evaluation of definite integrals

Question

If $J_{m} = \int_{1}^{e} \log^{m} x d x$ then $J_{8} + 8 J_{7}$ is equal to

Easy

A

2
B

$e^{8}$
C

$2 e$
D

$e$

Solution

$\begin{array}{l} J_{m} = \int_{1}^{e} \log^{m} x d x = {x \log^{m} x|}_{1}^{e} - \int_{1}^{e} m x \frac{\log^{m - 1} x}{x} d x \\ = e - m J_{m - 1} \end{array}$
So $J_{8} + 8 J_{7} = e$

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