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(D) Given that the area bounded by the curve $y = f(x)$,the $x-$ axis,and the lines $x = 1$ and $x = b$ is $\int_1^b f(x) \, dx = \sqrt{b^2 + 1} - \sq

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$\frac{x}{\sqrt{1 + x^2}}$

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(D) Given that the area bounded by the curve $y = f(x)$,the $x-$ axis,and the lines $x = 1$ and $x = b$ is $\int_1^b f(x) \, dx = \sqrt{b^2 + 1} - \sqrt{2}$.We can rewrite the expression as $\int_1^b f(x) \, dx = \sqrt{b^2 + 1} - \sqrt{1^2 + 1}$.This is equivalent to $\int_1^b f(x) \, dx = [\sqrt{x^2 + 1}]_1^b$.By the Fundamental Theorem of Calculus,if $\int_1^b f(x) \, dx = F(b) - F(1)$,then $f(x) = F'(x)$.Here,$F(x) = \sqrt{x^2 + 1}$.Therefore,$f(x) = \frac{d}{dx}(\sqrt{x^2 + 1})$.Using the chain rule,$f(x) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 1}}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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

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