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
Answer & explanation
Correct answer: option 4
We have,
$\int\limits_1^bf(x)dx=\sqrt{b^2+1}-\sqrt{2}$
Differentiating w.r. to b, we get
$f (b) =\frac{b}{\sqrt{b^2+1}}⇒f(x)=\frac{x}{\sqrt{x^2+1}}$