Definite Integration Question 217
Question: $ \int_a^{b}{\frac{\log x}{x}dx=} $
[MP PET 1994]
Options:
A) $ \log ( \frac{\log b}{\log a} ) $
B) $ \log (ab)\log ( \frac{b}{a} ) $
C) $ \frac{1}{2}\log (ab)\log ( \frac{b}{a} ) $
D) $ \frac{1}{2}\log (ab)\log ( \frac{a}{b} ) $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ I=\int_a^{b}{\frac{1}{x}\log xdx=(\log x\log x)_a^{b}} $
$ -\int_a^{b}{\frac{1}{x}\log xdx} $
$ \Rightarrow 2I=[{{(\log x)}^{2}}]_a^{b}\Rightarrow I=\frac{1}{2}[{{(\log b)}^{2}}-{{(\log a)}^{2}}] $
$ =\frac{1}{2}[(\log b+\log a)(\log b-\log a)] $ = $ \frac{1}{2}\log (ab)\log ( \frac{b}{a} ) $ .
BETA
