Does the second fundamental theorem of calculus easily imply the first fundamental theorem of calculus? The first does not imply the second because the second assumes only Riemann integrability of $f$ rather than continuity. The second only seems to imply the first if it is known that a continuous function $f$ possesses an antiderivative, which as far as I am aware is not a fact that is known prior to the fundamental theorem.
However, the Wikipedia page https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements makes the claim that the second fundamental theorem is "somewhat stronger" than the first. Am I correct that the second fundamental theorem does not easily imply the first?
Edit: As Miguel pointed out, the Wikipedia page claims that the second fundamental theorem is stronger than the corollary of the first fundamental theorem, not the first itself.
Best Answer
Everything you wanted to ask about the fundamental theorem but were afraid to.
On various forum questions I see way too many misconceptions about the relation between derivatives and integrals from current students and from former students. StackExchange does a good moderating job of clearing the crap. Take a look at similar questions and "stupid answers" on Quora for a real entertainment.
Q1. "If $F'(x)= f(x)$ for all $a\leq x \leq b$ then surely $\int_a^b f(x)\,dx =F(b)-F(a)$."
Sorry no, unless you learned your calculus in the 18th century and came here in a time machine. Calculus students now learn the Riemann integral. So this is false in general without assuming more about the function $f$. "But every calculus questions I ever solve uses antiderivatives." Well yes, but this is theory.
Yes, if you know that $f$ is continuous. We say $F$ is differentiable and $f$ is a derivative. So continuous functions are derivatives and Riemann integrable.
Yes, if you know that $f$ is Riemann integrable. Derivatives need not be Riemann integrable but some are Riemann integrable even if they are badly discontinuous.
A derivative does not have to be bounded, but Riemann integrable functions must be bounded.
"Can't I use the improper Riemann integral for unbounded derivatives?" Sure sometimes. But a derivative does not have to be integrable in that sense.
"Can I use the Lebesgue integral?" No. Not without further assumptions. Bounded derivatives are Lebesgue integrable, but unbounded ones need not be.
"How can I know that a function is or is not a derivative?" Well continuous functions are always derivatives. And derivatives have no jump discontinuities. But the question is beyond your pay-grade as a calculus student except for these two simple facts.
Q2. "If $f$ is Riemann integrable on $[a,b]$ and $F(x)=\int_a^x f(t)\,dt$ then isn't $F$ necessarily an antiderivative for $f$?"
If $f$ is continuous, yes.
If $f$ is continuous at one particular point $x_0$ then, indeed, $F'(x_0)=f(x_0)$, but it can fail at points where $f$ is not continuous.
So $F'(x_0)=f(x_0)$ if and only if $f$ is continuous at $x_0$. No. There can be other points.
P.S. "Easily" [See poster's question.]
Suppose I prove that (i) $f$ is Riemann integrable on $[a,b]$, and (ii) $F'=f$ everywhere, then $\int_a^b f(x)\,dx=F(b)-F(a)$. Does it follow easily that I can replace (i) by "$f$ is continuous on $[a,b]$"? No, you still have to prove that continuous functions are integrable.
But that is a feeble version of the FTC anyway. Try this one:
Theorem. Suppose that $f$ is Riemann integrable on $[a,b]$ and that there is a continuous function $F$ so that $F'(x)=f(x)$ with countably many exceptions. Then $\int_a^b f(x)\,dx=F(b)-F(a)$.