Definite Integration Question 416
Question: If $ g(1)=g(2) $ , then $ \int_1^{2}{{{[ fg(x) ]}^{-1}}}f’{g(x)}\ g’(x)\ dx $ is equal to
[AMU 2005]
Options:
A) 1
B) 2
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_1^{2}{{{[f{g(x)}]}^{-1}}f’[g(x)]g’(x)\ dx} $
Let $ f{g(x)}=z $
therefore $ f’{g(x)}\ g’(x)\ dx=dz $
When $ x=1,\ z=f\ {g\ (1)} $
When $ x=2,\ z=f\ {g\ (2)} $
$ \therefore $ $ I=\int _{f{g(1)}}^{f{g(2)}}{\frac{1}{z}dz}=|\log z| _{f{g(1)}}^{f{g(2)}} $
$ \Rightarrow $ I $ =\log f\ {g(2)}-\log f\ {g(1)}=0 $ , ( $ \because $ g(2)=g(1)).
BETA
