Definite Integration Question 438
Question: The area bounded by the x-axis, the curve $ y=f(x) $ and the lines $ x=1,x=b $ is equal to $ \sqrt{b^{2}+1}-\sqrt{2} $ for all b > 1, then $ f(x) $ is
[MP PET 2000; AMU 2000]
Options:
A) $ \sqrt{x-1} $
B) $ \sqrt{x+1} $
C) $ \sqrt{x^{2}+1} $
D) $ \frac{x}{\sqrt{1+x^{2}}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_1^{b}{f(x)dx=\sqrt{b^{2}+1}-\sqrt{2}}=\sqrt{b^{2}+1}-\sqrt{1+1}=[\sqrt{x^{2}+1}]_1^{b} $
$ \therefore $ $ f(x)=\frac{d}{dx}\sqrt{x^{2}+1} $
$ =\frac{2x}{2\sqrt{x^{2}+1}}=\frac{x}{\sqrt{x^{2}+1}} $ .
BETA
