Find $ F''(1)$ if
$ F(x) = \int_1^x f(t)dt $ $ f(t) = \int_1^{2t} \sqrt{1+u^3} du $
My work:
From the looks of it, it looks like the Fundamental Theorem of Calculus, twice.
From FTC... $F'(x) = f(x) $
On the second equation, derive both sides and use chain rule:
$f'(t)= d/dt*[ \int_1^{2t} \sqrt{1+u^3} du ] $
Using $z=2t$
$f'(t)=d/dz*[\int_1^{z} \sqrt{1+u^3} du]*dz/dt $
$f'(t) = 2 * \sqrt{1+(2t)^3} $
Just by gut feeling, I want to plug in $t=1$ and solve. Does $f'(t)$ represent $F''(x)$ ? $t$ and $x$ are different variable. Am I plugging in $1$ too early?