find a function f and a number a such that 6+ integral of t=x to t=a f(t)/t^2dt = 2square root x for all x>0.
Re-arrange: $\displaystyle \int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x}$.
Differentiate both sides wrt x: $\displaystyle \frac{f(x)}{x^2} = -\frac{1}{\sqrt{x}} \Rightarrow f(x) = -x^{3/2}$.
Substitute this expression into $\displaystyle \int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x}$ and solve for a.
I do realise that.
Are you familiar with the following property:
$\displaystyle \int_{\alpha}^{\beta} f(x) \, dx = - \int^{\alpha}_{\beta} f(x) \, dx$.
That's the property I used.
If you're familiar with the Fundamental Theorem of Calculus (see post #2) then you will understand why the integral terminals in your question need to be reversed before the integral can be differentiated.