Categories > definite-integral > Area bounded by region > 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

definite-integral

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


1) $\sqrt{x-1}$
2) $\sqrt{x+1}$
3) $\sqrt{{{x}^{2}}+1}$
4) $\frac{x}{\sqrt{1+{{x}^{2}}}}$
Solution: Explanation: No Explanation
Area bounded by region Volume and surface area of solids of revolution

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