Definite Integration Question 473
Question: If $ 2f(x)-3f( \frac{1}{x} )=x $ , then $ \int_1^{2}{f(x)}\ dx $ is equal to
[J & K 2005]
Options:
A) $ \frac{3}{5}\ln 2 $
B) $ \frac{-3}{5}(1+\ln 2) $
C) $ \frac{-3}{5}\ln 2 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2f(x)-3f( \frac{1}{x} )=x $ –(i)
Replacing x by $ ( \frac{1}{x} ) $ in (i), we get $ 2f( \frac{1}{x} )-3f(x)=\frac{1}{x} $ –.(ii)
Eliminating $ f( \frac{1}{x} ) $ from (i) and (ii), we get
$ -5\ f(x)=2x+\frac{3}{x}=\frac{2x^{2}+3}{3} $
therefore $ f(x)=-( \frac{2x^{2}+3}{5x} ) $
$ \int_1^{2}{f(x)dx=} $
$ -\int_1^{2}{( \frac{2x^{2}+3}{5x} )}\ dx=-\frac{1}{5}[x^{2}+3{\log _{e}}x]_1^{2} $
$ =-\frac{3}{5}[1+{\log _{e}}2]=-\frac{3}{5}[ 1+\ln 2 ] $ .
BETA
