The area bounded by the $x$-axis, the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $(\sqrt{\left( b^{2} + 1 \right)} - \sqrt{2})$ for all $b 1$, then $f(x)$ is

Question

The area bounded by the $x$-axis, the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $(\sqrt{\left( b^{2} + 1 \right)} - \sqrt{2})$ for all $b > 1$, then $f(x)$ is

Dr. Abubacker · Accepted Answer

(d)
$\int_{1}^{b}{f(x)dx = \sqrt{(b^{2} + 1)} - \sqrt{2}}$
On differentiating both sides $w.r.t.b$, we get
$f(b) = \frac{b}{\sqrt{(b^{2} + 1)}}$
Hence, $f(x) = \frac{x}{\sqrt{(x^{2} + 1)}}$

Q. The area bounded by the $x$-axis, the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $(\sqrt{\left( b^{2} + 1 \right)} - \sqrt{2})$ for all $b > 1$, then $f(x)$ is

Explanation

Reviewed by Dr. Abubacker

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