The area bounded by the $$x$$-axis, the curve $$y = f\left( x \right)$$ and the lines $$x =1,\,x = b,$$ is equal to $$\sqrt {{b^2} + 1} - \sqrt 2 $$ for all $$b > 1,$$ then $$f\left( x \right)$$ is :
A.
$$\sqrt {x - 1} $$
B.
$$\sqrt {x + 1} $$
C.
$$\sqrt {{x^2} + 1} $$
D.
$$\frac{x}{{\sqrt {1 + {x^2}} }}$$
Answer :
$$\frac{x}{{\sqrt {1 + {x^2}} }}$$
Solution :
Given $$\int\limits_1^b {f\left( x \right)dx} = \sqrt {{b^2} + 1} - \sqrt 2 $$
Differentiate with respect to $$b$$
$$f\left( b \right) = \frac{b}{{\sqrt {{b^2} + 1} }} \Rightarrow f\left( x \right) = \frac{x}{{\sqrt {{x^2} + 1} }}$$
Releted MCQ Question on Calculus >> Application of Integration
Releted Question 1
The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-