Some context: I'm an engineer, and I tend to have a rather unusual way of understanding and thinking about things, most likely related to my being autistic. I found this question on the HNQ and upon reading it I felt rather confused.

I have always understood that the indefinite and definite integrals are two sides of the same underlying concept, that there is no meaningful difference between them other than that one evaluates to a function and the other to a quantity. Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and $a$ a constant.

To me, whether the definite and indefinite integrals are two sides of the same thing, or two closely related but different things, seems more like a question of philosophy than mathematics.

Where is the flaw in my understanding? Is there one?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Sep 12, 2019 at 18:52

5 Answers 5


The point is, there are three slightly different concepts here:

1) If $f$ and $F$ are functions with the same domain, and $F'(x)=f(x)$ on that domain, then we say that $F$ is an antiderivative of $f$.

2) If $f$ is a function on the interval $a \le x \le b$, then the definite integral of $f$, $\int_a^b f(x) \, dx$, is (loosely speaking) the area under the graph of $y=f(x)$ for $a \le x \le b$, or (more precisely speaking) a limit of Riemann sums. This definition does not involve any derivatives.

3) If $f$ is a function on the interval $a < x < b$ and $c$ is a point in that interval, then the function $G(x)=\int_c^x f(t) \, dt$ could be called an indefinite integral of $f$. This definition of the function $G$ comes directly from the definition in 2) of the definite integral, so it also does not involve any derivatives.

You are saying that you don't see any difference between 2) and 3). You are correct! If your mental definition of "indefinite integral" is 3), then an indefinite integral is just a definite integral with some unknown limits.

The question you're linking is saying that many calculus books use 1) as the definition of an "indefinite integral": that is, they say that if $F$ is an antiderivative of $f$, then $\int f(x) \, dx = F(x)+C$ is the indefinite integral of $f$. They can get away with this because of the fundamental theorem of calculus, which says (in part) that

$$ \frac{d}{dx}\int_c^x f(t) \, dt=f(x) \, . $$

That is, the function $G(x)$ defined in 3) is an antiderivative of the function $f$, so these two notions of "indefinite integral" are closely related.

But closely related does not mean identical! K B Dave's answer is giving you some examples of situations where the two definitions 1) and 3) diverge slightly. Specifically:

  • We gave the definition 3) on an interval $a < x < b$, but it doesn't work very well on a more complicated domain. This means that definition 1) is more robust when $f$ has singularities.
  • The fundamental theorem of calculus says that every "indefinite integral" in the sense of 3) is an antiderivative. It does not say that every antiderivative is an indefinite integral, which in fact is false; you can have functions $f$ and $F$ where $F'(x)=f(x)$, but there is no way to write $F(x)=\int_c^x f(t) \, dt$.
  • 1
    $\begingroup$ Okay, it looks like what I fundamentally wasn't getting about the question that prompted this whole thing in the first place was that it was conflating "antiderivative" and "indefinite integral" to mean the same thing. My mental model is that antiderivative means 1), definite integral means 2), and indefinite integral means 3), and what I was thinking was that the question I linked (and people in the comments and other answers to this question) were saying that 2) and 3) were different things when they were saying that 1) and 2) were different things, which I understand better. $\endgroup$
    – Hearth
    Sep 9, 2019 at 16:13
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    $\begingroup$ Part of the problem might also have been that, in my field, 1) and 3) are often considered the same thing too, though there are minor differences. Engineers working with the real world don't often have to consider singularities, and when we do, we can fudge them by abusing delta functions and everything works out because the real world is full of low-pass filters. $\endgroup$
    – Hearth
    Sep 9, 2019 at 16:14
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    $\begingroup$ Yeah, on some level the point is that conflating 1) and 3) is computationally extremely handy, but conceptually confusing. As an engineer, you're more likely to care about the first half of that than the second; as a math educator, the person asking the other question is in the opposite situation. $\endgroup$
    – Micah
    Sep 9, 2019 at 16:17
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    $\begingroup$ Not all indefinite integrals are expressible as $$\int_c^x f(t)\, dt$$ The general form is actually $$\int_c^x f(t)\, dt + C$$ for some constant $C$. To understand the difference, note that the first expression must be $0$ for some $x$. Depending on the nature of $f$, the second expression need not be $0$ anywhere. Both expressions are antiderivatives of regular $f$, but only the second expression can represent all antiderivatives. $\endgroup$ Sep 9, 2019 at 23:53
  • $\begingroup$ Another point worth stressing: if the function you started with is not continuous, then FTC does not apply. Granted, provided that it is at least locally integrable, it still holds almost everywhere. $\endgroup$
    – tomasz
    Sep 10, 2019 at 10:54


"Reasonable" functions $f$ can have antiderivatives $D^{-1}f$ that are not of the form $x\mapsto \int_{a}^xf(t)\mathrm{d}t$ for any $a$ in the domain of $f$.

  1. Let $U=(-\infty,0)\cup(0,\infty)$, and consider $$\begin{align} f&: U\to\mathbb{R} & x&\mapsto \frac{1}{x}\text{.} \end{align}$$ Then $$D^{-1}f(x)=\begin{cases} \ln(-x)+C_- & x < 0\\ \ln(x) + C_+ & x > 0 \end{cases}$$ where $C_-$ and $C_+$ can differ—because the domain is not path-connected, there is no way to integrate from positive $x$ to negative $x$.
  2. Consider $$\begin{align} f&:\mathbb{R}\to \mathbb{R} & x&\mapsto \frac{1}{1+x^2}\text{.} \end{align}$$ Then $$\int_a^xf(t)\mathrm{d}t=\arctan x - \arctan a$$ but $$D^{-1}f(x)=\arctan x + C\text{;}$$ since the magnitude of $\arctan a$ is bounded by $\pi/2$, if $\lvert C \rvert \geq \pi/2$ then $D^{-1}f(x)$ is not of the form $\int_a^xf(t)\mathrm{d}t$.
  • $\begingroup$ I think this might answer the question, but I also think I'm going to have to brush up on my mathematical terminology to understand it. $\endgroup$
    – Hearth
    Sep 9, 2019 at 15:25
  • $\begingroup$ Yeah, my head hurts now. I think I'm going to have to come back to this later. $\endgroup$
    – Hearth
    Sep 9, 2019 at 15:50
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    $\begingroup$ @Hearth: I noticed your comments in the chat. In fact, you are right, and Michael Bacthold is totally wrong. He arrogantly insinuates that mathematicians cannot help you. While some mathematical notations are troublesome to formalize in a first-order theory, there is no trouble formalizing them in a more flexible system. Any proper logician can assure you of this, and you can read a sketch of one possible way to do it here. Feel free to clarify any details in chat. $\endgroup$
    – user21820
    Apr 16, 2020 at 10:59

As defined, the definite integral and the indefinite integral are totally different things.

The definite integral $\int_a^b f(x)\ dx$ is defined by looking at approximations to $f$ which are piecewise linear functions, looking at the areas of the resulting trapezoids, and then examining what happens in the limit as these approximations get more and more accurate.

The indefinite integral $\int f(x)\ dx$ is defined by searching for a function $g$ such that $g' = f$.

As you can see, on the face of it, these two concepts appear to have nothing to do with each other whatsoever. The definite integral has nothing to do with derivatives. And the indefinite integral has nothing to do with piecewise linear approximations or areas of trapezoids.

It just so happens that the definite integral and the indefinite integral are actually closely related: given some reasonable assumptions, if $F(x) = \int f(x)\ dx$, then $\int_a^b f(x)\ dx = F(b) - F(a)$. So the definite integral can, in fact, be found by finding the indefinite integral and then plugging some numbers into it. Likewise, the indefinite integral can be found by finding definite integrals where one endpoint is a variable. But the definitions are totally different.

Ultimately, maybe it doesn't matter. The Dedekind real numbers and the Cauchy real numbers are defined in completely different ways, but they end up defining exactly the same thing (at least in ZFC). Definite integrals and indefinite integrals aren't exactly the same, but unless you're doing real analysis on "ill-behaved" functions, the distinction doesn't really have any consequences.

  • 1
    $\begingroup$ It's not common to use trapezoids in defining the definite integral. Usually the definition is in terms of rectangles, as in the Riemann integral. But trapezoids can be useful in improving the accuracy of a numerical approximation of a definite integral. $\endgroup$
    – aschepler
    Sep 10, 2019 at 7:02

Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated [...No-one would claim they are "unrelated"; merely that the are "different". "different" most certainly does not mean "unrelated"...] to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and a $a$ constant.

I think this is your flaw. That $x$ is a variable and $a$ is a constant is a HUGE and very fundamental difference.

The number $0.86602540378443864676372317075294...$ can be thought of in many ways but ultimately it is "what it is". One of the things that it is is one half of the positive number that when squared is $3$. Another thing it is is $\sin \frac \pi 3$. The number doesn't change what it is just because we describe it in different ways for different purpose.

If $x$ is a variable then $\sin(x)$ is not a specific number (which $\sin a$ is). $\sin(x)$ is a theoretical possible unknown value of a number that would be sine of whatever number $x$ would be if it were pinned down. But as it isn't pinned down, it is an indefinite value. In a comment you claimed we don't call $\sin a$ the "definite sine" and $\sin x$ the "indefinite sine". We don't use those exact words but we most certainly do use those concepts and distinguish between them.

To add to the confusion the function $\sin$, itself, is different than $\sin(x)$. $\sin = \{(x,y)$ ordered pairs so that $y = \sin x\}$ is the concept of the function itself; a collection of all ordered pairs, whereas $\sin(x)$ is an indefinite output value of the function for an indefinite input value, and whereas $\sin (a)$ is a definite output value of the function for a definite input value.

These are three different applications of one concept or "three sides of the same coin". No-one is saying they are unrelated. But they are different things.


By the way, you've heard the joke about the a coworker who thinks the capital of Norway is in Sweden [a fundamentally impossible and inconsistent error; a capital of a country can't be in a different country so such an error is inconceivable for someone to make]? That's because he thinks Oslo is in Sweden [a stupid, but conceivable error] and Oslo is the capital of Norway [a fact; albeit one he is unaware of] and therefore he thinks that the capital of Norway [Oslo] is in Sweden.

This is kind of the same thing. The definite actuality of Oslo, and the indefinite concept of the capitol of a country are two different things even though they may evaluate to the same thing (if the country you choose is specifically Norway-- but not Sweden).

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    $\begingroup$ This is all good, except for the small problem that modern mathematics makes no distinction between a variable and a constant. Or could you give the definition of a variable element of $\mathbb{R}$ vs. a constant element of $\mathbb{R}$? See also the answer of Mike Shulman here $\endgroup$ Sep 10, 2019 at 13:50

(there are several ways to define integrals, and they are not 100% equivalent, so I'll go the elementary way to try to be clear; for the same reason, I will only discuss continuous functions)

We define the integral of the continuous function $f$ on the interval $[a,b]$ as $$\tag1 \int_a^b f(t)\,dt=\lim_{n\to\infty}\sum_{j=1}^n f(x_j)\,\Delta_j, $$ where $x_j=\frac{j(b-a)}n$ and $\Delta_j=1/n$ (more generally, one uses arbitrary partitions, but it is the same for continuous funcions).

If you think about it, you will note that there is no trace of a derivative nor anti-derivative in $(1)$. An integral is an integral, it has a million applications, and if $f$ is nice enough it can be approximated very well by taking $n$ big enough on the right-hand-side of $(1)$ (and maybe tweaking the $x_j$, like using the mid-point rule).

The relation with derivatives comes from Newton's wonderful Fundamental Theorem of Calculus: if $f$ is a continuous function, then the function $$\tag2 F(x)=\int_0^x f(t)\,dt $$ satisfies $F'(x)=f(x)$. Because of $(2)$, the notation $\int f(t)\,dt$ for an antiderivative of $f$ is so widespread. The fact that any two antiderivatives of $f$ differ by a constant leads to Barrow's Rule: $$\tag3 \int_a^b f(t)\,dt=F(b)-F(a). $$ So $(3)$ is the reason why we use antiderivatives to calculate integrals (when possible).

To emphasize the fact that integral means $(1)$, and not $(2)$ nor $(3)$, consider the Error Function $$ \operatorname{erf}(x)=\frac1{\sqrt\pi}\int_{-x}^x e^{-t^2}\,dt, $$ which is one of many important functions defined by an integral. The function $\operatorname{erf}(x)$ is a multiple of an anti-derivative of $f(t)=e^{-t^2}$, and no explicit expression for it exists.


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