# Finding a function (f) and a number a that satisfy the definite integral

Assume f is continuous. Then the derivative of $\int_a^x {f(t)\over t^2}dt$ is ${f(x)\over x^2}$, which must then be ${1\over \sqrt{x}}$ -- I took the derivative of both sides of the equation. So $f(x)={x^2\over \sqrt{x}}=x^{3/2}$. Now $\int_a^x {t^{3/2}\over t^2}dt=\int_a^x t^{-1/2}dt=2t^{1/2}|_a^x=2\sqrt{x}-2\sqrt{a}$. Finally then $6+2\sqrt{x}-2\sqrt{a}=2\sqrt{x}$ implies $3=\sqrt{a}$ or $a=9$.