Finding lower bound of integral

[susume]

"Find a function f and a number a such that
6 + integral{a to x} f(t)/t^2 dt = 2 sqrt(x)
for all x > 0."

I found f(x) but have no idea how to find a. Here's what I've done so far:

let g(x) = integral{a to x} f(t)/t^2 dt = 2 sqrt(x) - 6
differentiate: g'(x) = f(x)/x^2 = 1/sqrt(x)
f(x) = x^2 / sqrt(x) = x^(3/2)
g'(x) = x^(-1/2)
find anti-derivative: g(x) = 1/2 * x^(-1/2 + 1) + c = sqrt(x) / 2 + c
integral{a to x} g'(x) dx = g(x) - g(a) = sqrt(x)/2 - sqrt(a)/2 = 2 sqrt(x) - 6

I could solve for a in terms of x, but a is supposed to be a number, not a function of x. Any help will be much appreciated!

 

[tkhunny]

After finding f(x), why did you go back to reconstruct g'(x) differently?

Just evaluate the definite integral.

\(\displaystyle g(x) = \int_{a}^{x}\frac{1}{\sqrt{t}}\;dt\)

Do NOT get a "+c" when you do this.

 

[susume]

AH! I was trying to evaluate by using an anti-derivative. Didn't occur to me to ask what happens when x=a.
g(x) = integral{a to x} 1/sqrt(t) dt
g(a) = integral{a to a} 1/sqrt(t) dt = 0
6 + g(a) = 2 sqrt(a)
6 + 0 = 2 sqrt(a)
a = 9
Thank you!