# Using Fundamental Theorem of Calculus “Twice”

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?

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I didn't check every detail, but it looks like you are heading in the right direction. And yes $F''(1) = f'(1)$. This follows from your first application of the FTC. –  Code-Guru Dec 7 '12 at 2:00
You are not plugging in $1$ too early; $x, t, u$ are defined as separate variables, but just as $F'(x)=f(x)$, we can similarly say $f'(x)=2\sqrt{1+8x^3}$ by FTC and chain rule.
It may seem odd, since $x$ and $t$ are defined to be distinct variables, but what happens when you replace $t$ with $x$ in the second line? The math is correct; the issue here is not getting jumbled up by the variable notation.