Question

The area bounded by the x-axis, the curves 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

• A. $\sqrt{x-1}$
• B. $\sqrt{x+1}$
• C. $\sqrt{x^2+1}$
• D. $\frac{x}{\sqrt{1+x^2}}$

Answer 1

The correct option is D $\frac{x}{\sqrt{1+x^2}}$

${\int }_1^{b}f\left(x\right)dx=\sqrt{b^2+1}-\sqrt{2}=\sqrt{b^2+1}-\sqrt{1+1}=[\sqrt{x^2+1}{]}_1^b$
$\therefore f\left(x\right)=\frac{d}{dx}\sqrt{x^2+1}=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}$.