Definite Integration Question 164
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
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:
[d] Given $ \int\limits_1^{b}{f(x)dx=\sqrt{b^{2}+1}}-\sqrt{2} $ Differentiate with respect to b $ f(b)=\frac{b}{\sqrt{b^{2}+1}}\Rightarrow f(x)=\frac{x}{\sqrt{x^{2}+1}} $
BETA
