Definite Integration Question 365
Question: Let $ I_1=\int_1^{2}{\frac{dx}{\sqrt{1+x^{2}}}} $ and $ I_2=\int_1^{2}{\frac{dx}{x}} $ then
[Pb. CET 2004]
Options:
A) $ I_1>I_2 $
B) $ I_2>I_1 $
C) $ I_1=I_2 $
D) $ I_1>2I_2 $
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ (1+x^{2})>x^{2},\forall x $ ; $ \sqrt{1+x^{2}}>x,\forall x\in (1,2) $
therefore $ \frac{1}{\sqrt{1+x^{2}}}<\frac{1}{x},\forall x\in (1,2) $
therefore $ \int_1^{2}{\frac{dx}{\sqrt{1+x^{2}}}<\int_1^{2}{\frac{dx}{x}}} $
therefore $ I_1<I_2 $
therefore $ I_2>I_1 $ .
BETA
