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Q.
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
Application of Integrals
Solution:
We have $\int\limits_{1}^{b} f\left(x\right)dx =\sqrt{b^{2}+1}-\sqrt{2}$
On differentiating w.r.t. to $b$, we get
$f\left(b\right)=\frac{2b}{2\sqrt{b^{2}}+1}$
$\Rightarrow f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}$