Answer
$f(x)=x^{3/2}$
$a=9 $
Work Step by Step
Given
$$6+\int_a^x \frac{f(t)}{t^2} d t=2 \sqrt{x}$$
Differentiate the both sides with respect to $x$
\begin{aligned} \frac{d}{dx}(6)+ \frac{d}{dx}\int_a^x \frac{f(t)}{t^2} d t&= \frac{d}{dx}2 \sqrt{x}\\
0+\frac{f(x)}{x^2}&=\frac{1}{\sqrt{x}}\\
f(x)&= \frac{x^2}{\sqrt{x}}= x^{3/2} \end{aligned}
To find the value of $a$
\begin{aligned}
6+\int_a^x \frac{ t^{3/2} }{t^2} d t&=2 \sqrt{x}\\
6+\int_a^x t^{-1/2}dt&= 2\sqrt{x}\\
6+2\sqrt{x}- 2\sqrt{a}&= 2\sqrt{x}\\
\sqrt{a}&=3 \\
a&=9
\end{aligned}
