# Using the first fundamental theorem of calculus vs the second

So I'm unsure about using the fundamental theorem of calculus. The theorem I have is as follows:

Suppose $f(x)$ is continuous on $[a,b]$

1. If $g(x) = \int_{a}^{x} f(t) \,\mathrm{d}t$, then $g'(x) = f(x)$.
2. $\int_{a}^{b} f(x) \,\mathrm{d}x = F(b) - F(a)$ where $F$ is any antiderivative of $f$.

I'll use an example to elaborate my problem.

Let $F(x) = \int_{1}^{2x} \cos(t^2) \,\mathrm{d}t$. I want to find $F'(x)$.

So from examples I have seen I think, then $F(2x) = \cos (x^2)$, so to get $F'(x)$ we use the chain rule and our answer is $2\cos(4x^2)$. I know the answer is correct but have I written everything right?

Anyway, I somewhat understand this, but what if the limits where $1$ and $2$ say, rather than $1$ and $2x$? Do we use the second part of the theorem then? Are we just using the first part of the theorem in the question above because there's the variable $x$ involved (in $2x)$? Or does the second part of the theorem just state generally how to evaluate the integral? I think part of the issue is that we learn to integrate first and then are presented with this theorem and it seems like it is just stating what we already know.

Thank you

Let $f(x)=\int_1^x\cos t^2\,\mathrm dt$. Then $F(x)=f(2x)$ and therefore, by the chain rule,$$F'(x)=2f'(2x)=2\cos\bigl((2x)^2\bigr)=2\cos(4x^2).$$
If the upper limit is $2$ and not $2x$, then $F$ is contant and therefore $F'\equiv0$.