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So, from my understanding there are two versions of this theorem:

Version one states that, if $\displaystyle F(x)= \int_a^xf(t)~dt$, then $$\frac{dF}{dx}=\frac{d}{dx}\left[\int_a^xf(t)~dt\right]=f(x)$$whereas the second version states that $$\int_a^bf(x)~dx=F(b)-F(a)$$what I'm hoping to establish is this: I know that I can use the second version of the theorem to explain the first version, since $$\frac{dF}{dx}=\frac{d}{dx}\left[F(x)-F(a)\right]$$ $$=\frac{d}{dx}F(x)-\frac{d}{dx}F(a)$$and since each term in $F(a)$ will be a constant, we have that $$\frac{dF}{dx}=f(x)$$and in this regard, I understand why the theorem tells us that every function $f$ that is continuous on $[a,b]$ has an anti-derivative (or indefinite integral, if you like), $F$. What I'm trying to figure out, however, is whether or not this is a legitimate way of explaining the theorem? Is it true that both of these "versions" of the theorem are considered the same theorem? And if so, doesn't this mean that it's illegitimate to use the second version of the theorem to evaluate the first?

joshuaheckroodt
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  • https://math.stackexchange.com/questions/182691/fundamental-theorem-of-calculus-problem?rq=1 – user Dec 28 '17 at 13:54
  • no, your first assertion is not true. We need that $f$ must be continuous, then $F(x)=\int_a^x f(t),\mathrm dt\implies F'(x)=f(x)$. – Masacroso Dec 28 '17 at 14:00
  • Whenever you study a theorem you must focus on the hypotheses as well as the conclusions. Both the parts of FTC are essentially same if the function being integrated is continuous. If the function is discontinuous then both parts and different and the theorem has a different formulation than the one you mention. – Paramanand Singh Dec 29 '17 at 03:14
  • https://math.stackexchange.com/a/1900844/72031 – Paramanand Singh Dec 29 '17 at 03:35

2 Answers2

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You have only stated vaguely the FTC without giving the appropriate assumptions.

(First fundamental theorem of calculus.) Let ${[a,b]}$ be a compact interval of positive length. Let ${f: [a,b] \rightarrow {\bf C}}$ be a continuous function, and let ${F: [a,b] \rightarrow {\bf C}}$ be the indefinite integral ${F(x) := \int_a^x f(t)\ dt}$. Then ${F}$ is differentiable on ${[a,b]}$, with derivative ${F'(x) = f(x)}$ for all ${x \in [a,b]}$. In particular, ${F}$ is continuously differentiable.

(Second fundamental theorem of calculus.) Let ${F: [a,b] \rightarrow {\bf R}}$ be a differentiable function, such that ${F'}$ is Riemann integrable. Then the Riemann integral ${\int_a^b F'(x)\ dx}$ of ${F'}$ is equal to ${F(b) - F(a)}$. In particular, we have ${\int_a^b F'(x)\ dx = F(b)-F(a)}$ whenever ${F}$ is continuously differentiable.

These two versions are not the same. The first tells you that any continuous function has an "anti-derivative". The second one tells you something about the definite integral of the derivative of a differentiable function. Moreover, note that these two theorem have different sets of assumptions.

Note also that the differentiability of $F$ is in the conclusion of (I) but the assumption in (II).

Remark.

In a more advanced real analysis course, you will see that the two versions of FTC still holds with much weaker assumptions (with a slight cost that one only has $F'(x)=f(x)$ for almost every $x\in[a,b]$ in the conclusion of FTC I). On the other hand, the proofs of the two versions are quite different. See for instance this set of excellent lecture notes by Terry Tao.

Community
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  • Where is the dirty downvote from? Hmm, certainly not for a mathematical reason, is it? –  Jan 31 '18 at 22:16
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Define integrals $\int_{[a,b]} f(t)\>dt$ as limits of Riemann sums in your favorite way. The two versions of the FTC are relating such integrals to the notion of derivative, which comes as a miracle. The two formulas are: $${d\over dx}\int_{[a,x]}f(t)\>dt=f(x)\ ,\tag{1}$$ $$\int_{[a,b]} F'(t)\>dt=F(b)-F(a)\ .\tag{2}$$ Both of them seem to tell us that "integration" and "differentiation" are in a way "inverse" processes. The essential difference between the two is the following: In $(1)$ we first integrate a given function $f$, and then take the derivative of this integral with respect to, e.g., the upper limit. The result is the originally given $f$. In $(2)$ we first differentiate a given function $F$ and then obtain it back through integration in the second step.

Of course one has to prove both versions, but it suffices to work hard for one of these proofs, and the other then is an easy consequence. Usually one begins with $(1)$, but one could also start with $(2)$.

Christian Blatter
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