Definite Integration Question 430
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 > l, 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] Area $ =\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{1}{2}\frac{2x}{\sqrt{x^{2}+1}}=\frac{x}{\sqrt{x^{2}+1}} $
BETA
