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The Fundamental Theorem of Calculus says the following:

Theorem. If $f$ is the derivative of $F$ at every point on $[a,b]$, then under suitable hypotheses we have that $$\int_{a}^{b} f(t) \ dt = F(b)-F(a)$$

Theorem. If $f$ is integrable on $[a,b]$, then under suitable hypotheses we have that $$\frac{d}{dx} \int_{a}^{x} f(t) \ dt = f(x)$$

I am trying to put myself in the shoes of Poisson, Cauchy and Riemann. The first theorem is basically saying that to find the area under a curve, we need to find any anti-derivative and evaluate it at the endpoints?

The second theorem is saying that we can view the integral as a function of $x$ and take its derivative to get $f(x)$.

Wasn't the goal of Poisson, Cauchy and Riemann to find the area under a curve? So they hypothesized the first theorem and then only later proposed the second theorem? Both theorems deal with finding the area under a curve (i.e. they are equivalent)?

Do these theorems still hold under other types of integration (i.e. the Lebesgue integral)?

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For the Lebesgue integral, the function f must be not just continuous, but absolutely continuous for the FTC to hold. – gary Jun 23 '11 at 23:20
@gary - Absolute continuity is only needed for the first part of the FTC above. $f\in L^1$ is enough to guarantee that the second part holds at every Lebesgue point and thus almost everywhere. – Corey Jun 23 '11 at 23:31
Did you not give any consideration to comments left for you earlier today? – amWhy Jun 24 '11 at 0:04
Damien: This is technically the 19th question you've asked in roughly 2 days time, with a few of them asking two, sometimes three unrelated questions. I asked you earlier e.g, e.g.… You never did answer my question earlier today regarding the title and author of your text, which you say has no solutions. We'd be more than happy to steer you to some supplemental sources, so you can work more independently? – amWhy Jun 24 '11 at 0:24
The problem is, Damien, is that if you don't start picking and choosing more carefully which questions to ask, and how often to ask, you may find that, like "crying wolf", you run the risk of people here starting to ignore your questions, even though they may very well be urgent, important, good questions. – amWhy Jun 24 '11 at 0:30
up vote 28 down vote accepted


The relationship between the definite integral and the total change of an accumulation function goes back to well before Poisson, Cauchy, or Riemann. There's a nice historical overview in a recent article by David M. Bressoud, Historical reflections on teaching the Fundamental Theorem of Integral Calculus published in the Monthly last February. You can find one version in Leibniz's work in 1693, where he writes:

"I shall now show that the general problem of quadratures can be reduced to the finding of a line that has a given law of tangency, that is, for which the sides of the characteristic triangle have a given mutual relation. Then I shall show how this line can be described by a motion that I have invented."

"The problem of quadratures" is the problem of finding areas. Leibniz's proof, which is entirely geometric (you can find it in Bressoud's article) follows from the understanding of areas and tangents as certain sums and differences. But it does not originate with Leibniz: Isaac Barrow gave a proof in his Lectiones geometricae (1670); and James Gregory gives one in his Geometriae pars universalis (1668). Gregory shows that finding the length of a curve is equivalent to finding the area under a related curve: he shows that there is a constant $c$, chosen depending on certain given ratios, the length of the curve $y=f(x)$ from $x=a$ to $x=b$ equals the area under the curve $y=c\sqrt{1+(f'(x))^2}$ (though of course, not expressed that way). He then deals with the converse: given $y=g(x)$ on $[a,b]$, finding a curve $y=u(x)$ so that the area under the $y=g(x)$ is equal to the length of $y=u(x)$. He proves that if $$u(x) = \frac{1}{c}\int_a^x z(t)\,dt$$ then $z/c$ describes the slope of the tangent to $u$; this "contains" the second FTC.

Even earlier, the first part of what Gregory did had been done by Hendrick van Heureat, published in 1659 in van Schooten's edition of Descartes' Geometry.

Newton, by contrast, gives a kind of "dinamic proof" of the FTC; it has its roots in Oresme's Tractatus de configurationibus qualitatum et motuum (1350), in which he shows that if you represent velocity by a curve, then the area under the curve corresponds to the distance traveled (that is, the integral of the derivative equals the total change of the function, the first part of the FTC).

So by the time Cauchy and Riemann gave their definitions of integrals, the FTC (both parts) was already on "the table"; they had the onus of showing that their definitions implied the FTC. So the FTC was already "visible" to them (just like it was to Lebesgue), they didn't need to hypothesize the first or second theorem, nor propose them. They only had to show that their definitions were such that the theorems held for their integrals. Much like Lebesgue needed to show that his definition of integral agreed with that of Riemann where they were both defined, but that didn't mean he had to come up with Riemann's definition of integral from scratch: it was already there, he just needed to show his definition did not change the old properties.

Yes, there are versions of the Fundamental Theorem of Calculus that hold for other types of integrals. A good resource is A Garden of Integrals, by Frank E. Burke. The following statements are taken from there.

The Cauchy Integral

The Cauchy definition of integral (from 1823) is the following:

Given a bounded function $f$ on $[a,b]$, divide $[a,b]$ into a finite number of contiguous subintervals $[x_{k-1},x_k]$, $a= x_0\lt x_1\lt\cdots\lt x_n=b$. The Cauchy sum of $f$ is $$\sum_{k=1}^n f(x_{k-1})(x_k-x_{k-1}).$$ (This is the equivalent of a left-hand sum evaluation in today's parlance). We say that $f$ is Cauchy-integrable on $[a,b]$ if and only if there exists a number $A$ such that for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for any partition $P$ of $[a,b]$ whose subintervals have length less than $\delta$, we have $$\left|\sum_{P} f(x_{k-1})(x_k-x_{k-1}) - A\right| \lt \epsilon.$$ Cauchy proved that continuous functions are Cauchy-integrable, though this does not exhaust the class. We have the following "FTC"s for the Cauchy integral:

FTC for the Cauchy Integral. If $F$ is a differentiable function on $[a,b]$, and $F'$ is continuous on $[a,b]$, then $F'$ is Cauchy integrable on $[a,b]$ and $$C\int_a^x F'(t)\,dt = F(x)-F(a)$$ for each $x$ in $[a,b]$.

Here, $C\int$ denotes the Cauchy integral.

FTC part 2 for the Cauchy Integral. If $f$ is a continuous function on the interval $[a,b]$, and we define a function $F$ on $[a,b]$ by $F(x) = C\int_a^xf(t)\,dt$, then $F$ is differentiable on $[a,b]$, $F' = f$ on $[a,b]$, and $F$ is absolutely continuous on $[a,b]$.

We also have a convergence theorem:

Convergence for Cauchy Integrable Functions. If $\{f_k\}$ is a sequence of continuous functions converging uniformly to $f$ on $[a,b]$, then $f$ is Cauchy integrable on $[a,b]$ and $C\int_a^bf(x)\,dx = \lim C\int_a^b f_k(x)\,dx$.

The Riemann Integral

The definition of the Riemann integral is the usual one. Lebesgue proved in 1902 that a bounded function on $[a,b]$ is Riemann integrable on $[a,b]$ if and only if it is continuous almost everywhere.

FTC for the Riemann Integral. If $F$ is a differentiable function on the interval $[a,b]$, and $F'$ is bounded and continuous almost everywhere on $[a,b]$, then $F'$ is Riemann integrable on $[a,b]$, and $$R\int_a^x F'(t)\,dt = F(x) - F(a)$$ for each $x$ in the interval $[a,b]$.

FTC part 2 for the Riemann Integral. Suppose $f$ is a bounded and continuous almost everywhere function on the interval $[a,b]$. Let $F$ on $[a,b]$ be defined by $F(x)=R\int_a^x f(t)\,dt$. Then $F$ is absolutely continuous on $[a,b]$; if $f$ is continuous at $x_0\in[a,b]$, then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$; and $F'=f$ almost everywhere.

Convergence for Riemann Integrable Functions. If $\{f_k\}$ is a sequence of Riemann integrable functions converging uniformly to $f$ on $[a,b]$, then $f$ is Riemann integrable and $R\int_a^b f(x)\,dx = \lim R\int_a^b f_k(x)\,dx$.

Riemann-Stieltjes Integral

Let $f$ and $\phi$ be two bounded functions on $[a,b]$. We say that $f$ is Riemann-Stieltjes integrable with respect to $\phi$ if and only if there exists a number $A$ such that for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that $$\left|\sum_{k=1}^n f(c_k)\bigl(\phi(x_k) - \phi(x_{k-1})\bigr) - A\right|\lt \epsilon$$ where $x_{k-1}\leq c_k\leq x_k$, for every partition $P$ of $[a,b]$ whose subintervals has length less than $\delta$. We write $$RS\int_a^b f(x)d\phi(x) = A.$$

FTC for Riemann Stieltjes Integrals. If $f$ is continuous and $\phi$ is differentiable, with $\phi'$ Riemann integrable on $[a,b]$, then $$RS\int_a^b f(x)\,d\phi(x) = R\int_a^b f(x)\phi'(x)\,dx.$$

Theorem. If $f$ and $\phi$ are bounded functions with no common discontinuities on $[a,b]$, and the Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists, then the Riemann-Stieltjes integral of $\phi$ with respect to $f$ exists, and $$RS\int_a^b\phi(x)\,df(x) = f(b)\phi(b) - f(a)\phi(a) - RS\int_a^bf(x)\,d\phi(x).$$

FTC part two for Riemann-Stieltjes Integrals. If $f$ is continuous on $[a,b]$ and $\phi$ is monotonic increasing on $[a,b]$, then $f$ is Riemann-Stieltjes integrable with respect to $\phi$. If we define $F$ on $[a,b]$ by $$F(x) = RS\int_a^x f(t)d\phi(t),$$ then $F$ is continuous at any point where $\phi$ is continuous; and $F$ is differentiable at each point where $\phi$ is differentiable (almost everywhere) and at such points, $F' = f\phi'$.

Convergence Theorem for Riemann-Stieltjes Integrals. If $\{f_k\}$ is a sequence of functions that converge uniformly to $f$ on $[a,b]$ and $\phi$ is monotone increasing on $[a,b]$, then the $f_k$ is Riemann-Stieltjes integrable with respect to $\phi$ for each $k$, $f$ is Riemann-Stieltjes integrable with respect to $\phi$, and $$RS\int_a^bf(x)\,d\phi(x) = \lim RS\int_a^b f_k(x)\,d\phi(x).$$

Lebesgue Integral

FTC for the Lebesgue Integral. If $F$ is differentiable, and the derivative $F'$ is bounded on $[a,b]$, then $F'$ is Lebesgue integrable on $[a,b]$ and $$\int_{[a,x]}F'\,d\mu = F(x)-F(a)$$ for $x$ in $[a,b]$.

Lebesgue's FTC. If $F$ is absolutely continuous on $[a,b]$, then $F'$ is Lebesgue integrable and $$\int_{[a,x]} F'\,d\mu = F(x)- F(a)$$ for $x$ in $[a,b]$.

FTC Part 2 for the Lebesgue Integral. If $f$ is Lebesgue integrable on $[a,b]$, and we define $F$ on $[a,b]$ by $F(x) = \int_{[a,x]}f\,d\mu$, then $F$ is absolutely continuous on $[a,b]$ and $F'=f$ almost everywhere on $[a,b]$.

Dominated Convergence Theorem. If $\{f_k\}$ is a sequence of Lebesgue integrable functions converging pointwise almost everywhere to $f$ on $[a,b]$, and $g$ is a Lebesgue integrable function such that $|f_k|\leq g$ on $[a,b]$, then $f$ is Lebesgue integrable on $[a,b]$ and $$\int_{[a,b]}f\,d\mu = \lim \int_{[a,b]} f_k\,d\mu.$$

Henstock-Kurzweil Integral

A function $\delta\colon [a,b]\to (0,\infty)$ is called a gauge on $[a,b]$. A tagged partition of $[a,b]$ is a finite collection of pointed intervals $(c_k,[x_{k-1},x_k])$, where $x_{k-1}\leq c_k\leq x_k$, $a=x_0\lt x_1\lt x_2\lt\cdots\lt x_n=b$. We say a tagged partition of $[a,b]$ is $\delta$ fine if $c_k-\delta(c_k) \lt x_{k-1}\leq c_k \leq x_k \lt c_k+\delta(c_k)$.

A function $f$ on $[a,b]$ is said to be Henstock-Kurzweil integrable on $[a,b]$ if there is a number $A$ such that for every $\epsilon\gt 0$ there exists a positive function $\delta_{\epsilon}\colon [a,b]\to (0,\infty)$ such that for any $\delta_{\epsilon}$-fine partition on $[a,b]$ with $c_{k}-\delta_{\epsilon}(c_k) \lt x_{k-1}\leq c_k \leq x_{k}\lt c_k+\delta_{\epsilon}(c_k)$, we have: $$\left|\sum_{k=1}^n f(c_k)(x_k-x_{k-1}) - A\right|\lt \epsilon.$$ In that case, we write $HK\int_a^b f(x)\,dx = A$.

FTC for the Henstock-Kurzweil Integral. If $F$ is continuous on $[a,b]$ and $F$ is differentiable on $[a,b]$ with at most a countable number of exceptional points, then $F'$ is Henstock-Kurzweil integrable on $[a,b]$, and $$HK\int_a^x F'(t)\,dt = F(x) - F(a)$$ for each $x$ in $[a,b]$.

FTC part two for the Henstock-Kurzweil Integral. If $f$ is Henstock-Kurtzweil integrable on $[a,b]$, and we define $F$ by $F(x) = HK\int_a^x f(t)\,dt$, then $F$ is continuous on $[a,b]$, $F'=f$ almost everywhere, and $f$ is Lebesgue measurable.

Dominated Convergence for Henstock-Kurzweil Integral. If $\{f_k\}$ is a sequence of Henstock-Kurzweil integrable functions that converge pointwise to $f$ on $[a,b]$, and there exist Henstock-Kurzweil integrable functions $\phi$ and $\psi$ such that $\phi\leq f_k\leq \psi$ for all $k$, then $f$ is Henstock-Kurzweil integrable and $$HK\int_a^b f(x)\,dx = \lim HK\int_a^b f_k(x)\,dx.$$

The integrals above are given in increasing order of strength, in the following sense: if $f$ is a function on $[a,b]$, then: $$\begin{align*} f\text{ is Cauchy integrable on }[a,b] &\Longrightarrow f\text{ is Riemann integrable on }[a,b]\\ &\Longrightarrow f\text{ is Lebesgue integrable on }[a,b]\\ &\Longrightarrow f\text{ is Henstock-Kurzweil integrable on }[a,b] \end{align*}$$ and none of the implications are reversible. The Riemann-Stieltjes (and its variant, the Lebesgue-Stieltjes) integral is not included in the chain of implications, because integrability there depends on both the function $f$ and the function $\phi$.

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As far as I can tell, this is the third fantastic answer of yours based on that book. I definitely need to take a look at it! See here and here to see what I mean. – t.b. Jun 24 '11 at 17:45
@Theo: Thanks for the kind words. The book is very interesting, and very readable. Highly recommended. – Arturo Magidin Jun 24 '11 at 18:46

I only have a partial answer.

First of all, try drawing some figures that could belong to those theorems, then they don't seem so mysterious at all.

If a function $f:[a,b] \to \mathbf R$ is Riemann integrable then it is Lebesgue integrable and the integrals coincide.

If $f:[a,b] \to \mathbf R$ is bounded then $f$ is Riemann integrable if and only if the points where $f$ is discontinuous on $(a,b)$ is a null set.

Conversely if we consider the second equation then according to the Lebesgue differentiation theorem we obtain $f(x)$ "almost everywhere".

This gives an answer to your last question.

Another comment is that it is for some problems fine to think about an integral as the area under the curve but for more "advanced" analysis this is not so suitable. I think of them as linear operators.

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That is a good addition. – Jonas Teuwen Jun 23 '11 at 23:24
Aaah, finally! Hard earned. – t.b. Jun 23 '11 at 23:34
You probably already know this but it is worth stating that even though one usually views the Lebesgue integral as an operator, one can also view the Lebesgue integral as "the area under the graph" (using Fubini's theorem, for example). – Amitesh Datta Jun 24 '11 at 2:02

The following answer might be somewhat sophisticated but it presents a rather intuitive proof of the Lebesgue differentiation theorem based on the weak type (1,1) property satisfied by the Hardy-Littlewood maximal operator. More specifically, the following result is known as Lebesgue's differentiation theorem:

Lebesgue's differentiation theorem: For any locally integrable function $f$ on $\mathbb{R}^n$ we have

$\lim_{r\to 0} \frac{1}{\left|B(x,r)\right|} \int_{B(x,r)} f(y)dy = f(x)$

for almost all $x\in \mathbb{R}^n$. Consequently we have $\left|f\right|\leq {\cal M}(f)$ a.e.

The idea is that if we can control the operator $T^{*}$ defined by the rule $T^{*}(f)=\sup_{r>0} \left|\int_{B(x,r)} f(y)dy\right|$ by an operator with nice boundedness properties (e.g., the Hardy-Littlewood maximal operator), then we can prove the differentiation theorem above. More precisely, we prove the differentiation theorem above for continuous functions $f$ with compact support (this is easy), and we then use the density of the space of compactly supported continuous functions in $L^1$.

Intuitively, the "trick" is that the boundedness properties of the operator $T^{*}$ imply a certain "bounded oscillation condition" that allows us to prove the result for all locally integrable functions using the validity of the result for compactly supported continuous functions. The more precise explanation can be found in Loukas Grafakos' "Classical and Modern Fourier Analysis", Chapter 2, Section 1, Theorem 2.1.14, page 86.

I like this proof of the differentiation theorem very much because it uses the weak type (1,1) property of the Hardy-Littlewood maximal operator and is based on a result that has various applications (including the solution of the Dirichlet problem in upper half space). Finally, it provides ample evidence for the importance of the Hardy-Littlewood maximal operator in harmonic analysis.

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I tried putting myself in the shoes of Leibniz. And here is my perception. Which I can only attempt to explain.

Consider the definition of the derivative

$$f'(x) = \lim_{h\to0}\dfrac{f(x + h) - f(x)}{h}$$ Let A represent the area of a rectangle whose base is x and height is f(x). (At this time it helps to draw a concave up curve in the 1st quadrant of a plane.) That is $$A=xf(x)$$ From the definition of the derivative

$$A'(x) = \lim_{h\to0}\dfrac{(x + h)f(x + h) - xf(x)}{h}$$

$$A'(x) = \lim_{h\to0}\dfrac{(x)f(x + h) - xf(x)+h(f(x)}{h} $$ $$A'(x) = {xf'(x) + f(x)}$$ It is worth noticing that the definition of A implies.

$$\dfrac{d(xf(x))}{dx}={xf'(x) + f(x)}$$

As for me, I would end it here. That is the relationship between the tangent and the area. But, for the rest of us, it is clear that A(x) can be written as the sum of the area under f(x), A1, and the area above and to the left of f(x), A2.
That is $A(x) = A1(x)+A2(x)$

$$\dfrac{dA2}{dx}+\dfrac{dA1}{dx}={xf'(x) + f(x)}$$

Since f(x) is arbitrary, then it follows that so long as f(x) is differentiable one, or both, of the following conditions should hold true:

Condition 1: $\dfrac{dA2}{dx}={xf'(x)}$ and $\dfrac{dA1}{dx}={f(x)}$

Condition 2: $\dfrac{dA1}{dx}={xf'(x)}$ and $\dfrac{dA2}{dx}={f(x)}$

After some trial and error with several forms of f(x) it seems that condition 1, always holds true. The limits of integration follow, these are a necessary condition to determine the A, A1, and A2 over a given interval.

So far so good. Now let $f(x)=1/G(x)$ in this case $A=x/G(x)$ using the definition of the derivative

$$A'(x) = \lim_{h\to0}[\dfrac{x+h}{G(x + h)} -\dfrac{x}{G(x)}]$$ $$A'(x) = \dfrac{1}{G(x)} - \dfrac {x}{(G(x))^2}\dfrac {dG(x)}{dx}$$

$$\dfrac{d(x/G(x))}{dx} = \dfrac{1}{G(x)} - \dfrac {x}{(G(x))^2}\dfrac {dG(x)}{dx}$$

Looks familiar...

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