What is wrong with the following proof of the Second Fundamental Theorem of Integral Calculus?

Question

An exercise in a textbook I am using to study integral calculus posits the following query: what is wrong with the following proof of the Second Fundamental Theorem of Integral Calculus? (The theorem being that: if $f$ is continuous on $[a,b]$, then $F(x) = \frac{d}{dx}\int_a^x f(t) dt$ is differentiable at every point $x$ in $[a,b]$ and $\frac{dF}{dt} = f(x)$).

Step 1. Let $G(x) = \int_a^x f(t) dt$

Step 2. Let $F$ be an antiderivative of $f$.

Step 3. Then, $G(x) = \int_a^x f(t) dt = F(x) – F(a)$

Step 4. Differentiate both sides of the equation in Step 3 to get $\frac{dG}{dx} = \frac{d}{dx}F(x) = f(x)$

I don't quite see why this proof is invalid. I assume the author assumed that $f(t)$ is continuous on $[a,b]$. Step 1 is just a construction; Step 2 is justified since all continuous functions have an antiderivative, hence $F$ can indeed be an antiderivative of $f$; Step 3 is justified with the First Fundamental Theorem of Integral Calculus, because $x$ is within $a$ and $b$; Step 4. is justified through basic derivative principles.

I suspect what the author might be aiming for is that the proof didn't really show that $G(x)$ is differentiable at every point $x$ in $[a,b]$. Is this what is meant to be pointed out by the exercise? Guidance would be much useful. Thank you in advance.

Answer 1

The problem with the proof is Step 2; specifically, Step 2 assumes that you know that a function that is continuous on a finite closed interval has an antiderivative... but that is precisely the content of the theorem that you are trying to prove.

The two parts of the Fundamental Theorem of Calculus say that if $f(x)$ is continuous on $[a,b]$, then:

1. Part 1: If $F(x)$ is an antiderivative of $f(x)$ on $[a,b]$, then $\int_a^b f(t)\,dt = F(b)-F(a)$; and
2. Part 2: The function $G(x) = \int_a^x f(t)\,dt$ is differentiable on $[a,b]$ and $\frac{d}{dx}G(x) = f(x)$ on $[a,b]$. That is, $G(x)$ is an antiderivative of $f(x)$ on $[a,b]$.

I like to think of them as saying:

Part 1 says that if you have an antiderivative of $f(x)$, then you can use the values of the antiderivative to compute $\int_a^b f(t)\,dt$.

And Part 2 tells you that an antiderivative of $f(x)$ exists, at least in principle.

Unfortunately, the antiderivative you get in Part 2 is not "useful" for Part 1 (if you try to use it to calculate the integral, you end up showing that the value of the integral is the value of the integral). But it does tell you that the quest to find an antiderivative of $f(x)$ is not doomed to fail: antiderivatives of continuous functions exist; and Part 1 tells you that if you happen to find one that you can calculate in some way that is independent of the calculation of the integral of $f(x)$, then you can use it to find the value of $\int_a^b f(t)\,dt$.

In the absence of Part 2, it is not at all clear that antiderivatives exist at all. So your argument is circular, in that it assumes that an antiderivative exists in order to prove that an antiderivative exists. If you happen to know that $f(x)$ has an antiderivative in some independent way, usually for specific functions, then your argument works; but for arbitrary functions the whole point of the FTC Part 2 is to establish that existence.