I want to understand the connection between the primitive function or antiderivative and the definite integral.

My problem with this is the independent variable called t in the formula for the first part of the Fundamental Theorem of Calculus.

Here's a composite of the answers I've already seen for this question. Because of t I don't understand it:

The primitive is a function $F(x)$ such that

$F′(x) = f(x)$

The first part of the FTC, which is the thing that connects differentiation/antidifferentiation to the definite integral, is this:

$$F(x) = \int_a^x f(t) dt $$ or “$F(x)$ is a primitive function of $f(x)$. The lower bounds a is fixed. The upper bounds $x$ is variable.”

You can write $\int f(x) dx$ as $$\int_a^x f(t) dt + C$$

A complete analysis of this problem is given in Richard Courant’s calculus book (linked below) page 109+.

Some of the sites I’ve seen: http://ia700700.us.archive.org/34/items/DifferentialIntegralCalculusVolI/Courant-DifferentialIntegralCalculusVolI.pdf


web.utk.edu/~wneilson/mathbook.pdf page 153-154






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    $\begingroup$ Too long; didn't read. What's the question? $\endgroup$ Feb 14, 2012 at 11:13
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    $\begingroup$ @GerryMyerson, +1, If you want quicker answers (& possibly, better answers). Summarize what you wish to ask. $\endgroup$
    – Inquest
    Feb 14, 2012 at 11:51
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    $\begingroup$ I also don't understand the question. The answer to the question asked in the title is simply "the fundamental theorem of calculus." Is that what your question is about? $\endgroup$ Feb 14, 2012 at 17:50
  • $\begingroup$ Let's discretize the situation and perhaps this will help you tell us what your real confusion is. One can replace functions on the real line by sequences $a_n, n \in \mathbb{Z}$. The analogue of differentiation is taking finite differences $b_n = a_{n+1} - a_n$. The analogue of definite integration is taking sums $\sum_{i=n}^m a_i$, and the analogue of indefinite integration is taking indefinite sums $b_n = \sum_{i=n_0}^n a_i$. The "fundamental theorem of finite differences" says the obvious thing and is straightforward to prove. What is confusing about this situation? $\endgroup$ Feb 14, 2012 at 17:53
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    $\begingroup$ Have you seen this answer? $\endgroup$ Feb 14, 2012 at 17:56

3 Answers 3


I'll try to be a succint as possible but answering your doubts as far as I can, but you made too many questions. Try to summarize as far as you can, adressing your main concerns that will clear the secondary ones.

A. I want to understand deep down intuitively the connection between the primitive function or antiderivative and the definite integral. I am already aware of at least this much:

The so-called indefinite integral is not an integral. Integrals can be represented as areas but the indefinite integral has no bounds so is not an area and therefore not an integral.

The indefinite integral, in my opinion, should be called "primitive" to avoid confusions, as many people call it. The idea is that we learn how to find derivatives, and then are told: "Well, but what about the inverse problem: If we have a known function, what function should we differentiate to get it?" And here comes the idea of primitive of a function, or rather, primitives. The primitive of a given function, which we denote by

$$\int{ f(x) }dx = F(x)$$

is a function such that $F'(x) = f(x) \text{ ; } (1)$.

The notation was used by Leibniz to denote a function that satisfied $(1)$, using the arbitrary constant $C$ to denote that the function wasn't unique - rather, there was a family of primitives of a given function $f$, since the derivative of a constant is null. This is simple notation, but it has nothing to do with $\int_a^b f$ in the sense that $\int_a^b f$ is a number representing the limit of an integral sum of $f$ over $I = (a,b)$, and $\displaystyle \int f(x) dx +C $ is representing a function. As you say, $$\int{ f(x) }dx + C = F(x)$$ is not an integral sum, but a symbolysm for the function that satisfies $(1)$.

B. The first part of the FTC, which is the thing that connects differentiation/antidifferentiation to the definite integral, is this: (...) etc.

Here you're getting confused. The FTC states:

Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by

$$F(x) = \int_a^x f(t) dt $$

Then $$F'(x) = f(x)$$

for all $x$ in $[a, b]$.

This, in short, says that $F$ is indeed another primitive of $f$.

A consequence of this is the so called second FTC and a corollary. Since two primitives are only different by a constant, we should have

$$F(x) - \int_a^x f(t) dt = C$$

But then putting, $x = a$ gives

$$F(a) = C$$

which states that

$$F(x) - \int_a^x f(t) dt = F(a)$$


$$F(x) - F(a) = \int_a^x f(t) dt$$

Plugging in $b$ as the upper bound gives the famous corollary:

$$F(b) - F(a) = \int_a^b f(t) dt$$

which states that if $F$ is a primitive of $f$, the previous equality holds.

The second FTC says:

Let $f$ be a function defined on a closed interval $[a, b]$ that admits a primitive $F$ on $[a, b]$, i.e.:

$$f(x) = F'(x)$$

If $f$ is integrable on $[a, b]$ then

$$\int_a^b f(x) dx = F(b) - F(a)$$

(And this didn't depend on the continuity of $f$! You can try and plot the integrals dependeing on the upper bound of discountinous functions to see how the integral always "behaves" much better than the integrand.)

The connection between the primitives and the deinite integral is thus: If we know that a function admits a primitive in an interval, we can easily calculate it's definite integral by means of the primitive.

  • $\begingroup$ I am having conceptual difficulty with the fact that $C=F(a)$. I am perfectly comfortable with the notion that $F(x)=\int f(x)dx+C$ is a family of primitives differing by a constant, as well as with the fact that the definite integral $\int_a^b f(x)dx=F(b)−F(a)$ is a limit of Riemann sums over the interval $[a,b]$ and is equal to the area bounded by $x=a,x=b,y=0$, and $y=f(x)$. However, the notion $\int_a^x f(t)dt$ is strange as it seems to be a hybrid of definite and indefinite integrals. I see from the FTC that $\int_a^x f(t)dt=F(x)−F(a)$, therefore $F(a)$ must equal $C$, (cont.) $\endgroup$
    – user001
    Mar 8, 2012 at 14:49
  • $\begingroup$ (cont.) but I do not follow the logic that since any two primitives differ by a constant, $F(x) - \int_a^x f(t)dt = F(a)$. Again, I see by using the FTC that $F(x) - \int_a^x f(t)dt = F(x) - [F(x) - F(a)] = F(a)$, but I am having trouble imagining what is $\int_a^x f(t)dt$. It seems to be neither an indefinite nor a definite integral. And how geometrically is it that by adding $F(a)$ to it, I recover $F(x)$ which is a family of primitives? I'm trying to visualize this, but when I envision $F(x)$, I see a family of vertically translated curves and when I picture $\int_a^x f(t)dt$, (cont. $\endgroup$
    – user001
    Mar 8, 2012 at 14:50
  • $\begingroup$ (cont.) I see a definite integral with a sliding upper bound, which is numerically just the area bounded by $x=a,y=0,y=f(x)$ and that sliding bound. I guess what is unclear is the relation between $C$ and the lower bound. How is that specifying a lower bound fixes $C$. Again, I can write out the math (as above) and see that they must be equal, but I'm lacking an intuitive understanding. $\endgroup$
    – user001
    Mar 8, 2012 at 14:52
  • $\begingroup$ @user001 We proved that $\int\limits_a^x f(t) dt$ is a primitive of $f(x)$. The upper limit, as you say, is variable, so this is a function that retrieves the area under $f$ in an interval $(a,x)$. I stated that two primitives differ by a constant. By putting $x=a$ we have that $F(a)-\int\limits_a^a f(t) dt=C$. Since the integral is $0$ we get that $F(a) =C$. That is what you're writing very comfortably: $\int\limits_a^b f(t) dt = F(b)-F(a)$. There is no need for intuition, it is an algebraic equality. In this case $F(x)$ is NOT a familiy of primitive but A primitive, \\...\\ $\endgroup$
    – Pedro
    Mar 8, 2012 at 15:03
  • $\begingroup$ \\...\\ so it is a single curve. Don't let geometry get in your way. Think analytically for a while. When I write $\int\limits_a^x f(t) dt$ I use $t$ as a dummy variable so as not to arise conflict with $x$. You could put $\int\limits_a^x f(x) dx$, but it will make things confusing. Think about $\int\limits_a^x f(t) dt$ as a function of $x$. Try and plot $\int\limits_0^x t dt$ for some values of $x$, and see what you get. $\endgroup$
    – Pedro
    Mar 8, 2012 at 15:06

The definite integral $$\int_a^b f(x)\,dx$$ is a number: specifically, it is the net signed area between the lines $x=a$, $x=b$, $y=0$, and the curve $y=f(x)$.

(It is defined to be a limit, if it exists, of certain finite sums, called Riemann sums; but the sums and limit are designed to try to capture the idea of area; see for example this previous answer).

Now, one can try to compute the integral as a limit of certain finite sums. What the Fundamental Theorem of Calculus, Part II, tells you, is that these computations are intimately (though surprisingly) connected with differentiation.

Let's start by considering a function defined on $[a,b]$, and that has an integral $\int_a^b f(x)\,dx$.

Now, for each number $c$, $a\lt c\lt b$, we can consider just the area between $x=a$ and $x=c$, instead of going all the way to $x=b$; this, of course, is $\int_a^c f(x)\,dx$. For each value of $c$, this is a number. So we have defined a new function: the "domain" of the function is "all numbers between $a$ and $b$", and the rule for the function is: "given $c$ in the domain, the value of the function is the net signed area between $x=a$, $x=c$, $y=0$, and $y=f(x)$." Though complicated to write out, this is a real-valued function of real variable: a rule that assigns to every valid input (numbers $c$ between $a$ and $b$) one and only one output (the net signed area corresponding to $\int_a^c f(x)\,dx$). That is, we have a function $$c\longmapsto \int_a^c f(x)\,dx.$$

Another way of writing this is to call this function $\mathcal{F}$, and call the input $t$ (since it's the variable, but $x$ is already in use), and write: $$\mathcal{F}(t) = \int_a^t f(x)\,dx$$ so that $\mathcal{F}(t)$ tells you the net signed area "up to $t$".

The Fundamental Theorem of Calculus, Part II, tells you that if you do this, and then ask "is $\mathcal{F}(t)$ differentiable?", the answer is "Yes, it is differentiable, and in fact: $$\frac{d}{dt}\mathcal{F}(t) = f(t)."$$ (See for example this answer for a graphical explanation).

This leads, in turn, to the Fundamental Theorem of Calculus, Part I, which says that there is an alternative way to compute the integral. Namely, if you can find any function $\mathcal{F}(x)$ with the property that $\frac{d}{dx}\mathcal{F}(x) = f(x)$ for all $x$ in $[a,b]$, then $$\int_a^b f(x)\,dx = \mathcal{F}(b)-\mathcal{F}(a).$$

That is, the First Fundamental Theorem of Calculus says: "There's a way you can avoid doing all those annoying finite sums and limits: if you can find a function whose derivative is $f$, then you can use that function to compute the integral."

Now, if we can find a function $\mathcal{F}(x)$ whose derivative is $f(x)$ for all $x$, then $\mathcal{F}(x)$ can be used to compute any definite integral of $f(x)$: just plug in the limits, get the integral. Since this is much simpler than trying to do the Riemann sums and the limits, though by no means a trivial task, we are led to the following: instead of trying to do each integral we are faced by doing limits of Riemann sums, we will instead try to do the following:

Given a continuous function $f(x)$, find a function $\mathcal{F}(x)$ such that $\frac{d}{dx}\mathcal{F}(x) = f(x)$ for all $x$.

We call the function $\mathcal{F}(x)$ an "antiderivative of $f(x)$".

So now, instead of "find the net signed area between $x=a$, $x=b$, $y=0$, and $y=f(x)$", we are faced with the task "find an antiderivative of $f(x)$."

(Note that while there is one and only one value for the area, there are many antiderivatives; however, the Constant Function Theorem tells us that if $\mathcal{F}(x)$ and $\mathcal{G}(x)$ are both antiderivatives of $f(x)$, then $\mathcal{F}(x) - \mathcal{G}(x)$ is constant; that means that if we can find one antiderivative $\mathcal{F}(x)$, then any other antiderivative will be of the form $\mathcal{F}(x)+C$, where $C$ is a constant).

Just like "the net signed area between $x=a$, $x=b$, $y=0$, and $y=f(x)$" is abbreviated with a convenient symbol, $$\int_a^b f(x)\,dx$$ (which has some nice notational properties, such as additivity, etc.), we also want some convenient symbol for

"the general antiderivative of the function $f(x)$."

Why? Because (i) we don't want to have to say "the general antiderivative of the function $f(x)$" all the time (we're lazy); (ii) because good notation is suggestive, and if we can find a good notation that has some nice notational properties, that will help in solving problems down the road.

Since "a general antiderivative of the function $f(x)$" (call it $\mathcal{F}(x)$) can be used to find any definite integral via the Fundamental Theorem of Calculus, connecting the notion to the definite integral is a good idea (though this is the main use of antiderivatives students encounter early on, they turn out to be important for other things, particularly to solve differential equations that turn up when we describe physical phenomena all the time). And so, since $\mathcal{F}(x)$ is useful to compute $$\int_{\square}^{\triangle} f(x)\,dx$$ for any $\square$ and $\triangle$, we leave those spaces "blank" to suggest "you can use this for any integral". This leads to the notation $$\int f(x)\,dx$$ for an antiderivative of $f(x)$. This leads to a bit of a problem, though: since there is no single answer (many different antiderivatives), we don't want the same symbol to represent different things. So we use the symbol above to represent any and all antiderivatives of $f(x)$ (or "the most general antiderivative of $f(x)$"). So when we write something like $$\int x\,dx = \frac{1}{2}x^2 + C$$ we are saying: "the antiderivatives of $f(x)=x$ are the functions $\mathcal{F}(x) = \frac{1}{2}x^2 + C$ with $C$ constant."

  • $\begingroup$ I'm certain that the FTC I is the fact that if $$\mathcal{F}(t) = \int_a^t f(x) dx$$ then $$\mathcal{F}'(t) = f(t)$$ and FTC II is $$\int_a^b f(x) dx = \mathcal{F}(b)-\mathcal{F}(a)$$ $\endgroup$
    – Pedro
    Feb 14, 2012 at 21:07
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    $\begingroup$ @Peter: I'm certain that it depends on your source. Stewart gives it as you do. Anton, Bivens, and Davis give it as I do (Theorems 6.6.1 and 6.6.3 in the 7th Edition of "Early Transcendentals"), as do Hughes-Hallett, McCallum, et al. $\endgroup$ Feb 14, 2012 at 21:17
  • $\begingroup$ I have answers...the question is about the part of the answer I don't understand, which is the variable t (Courant used u). Why it's there, what it represents. From my question part A: "My biggest problem with this is probably the switching to an independent variable called t... ...Here is that answer, and I don’t understand all of it: The primitive is a function F(x) such that F′(x)=f(x) The first part of the FTC, which is the thing that connects differentiation/antidifferentiation to the definite integral, is this: F(x)=∫f(t)dt[ax]" $\endgroup$
    – Luther
    Feb 15, 2012 at 6:13
  • $\begingroup$ @Luther: The $t$ is the variable in the function $\mathcal{F}$. The function $\mathcal{F}$ is defined to be: given an input $t$, the output is the net signed area between $x=a$, $x=t$, $y=0$, and $y=f(x)$." I wrote that very clearly in the post. That's the definition of $\mathcal{F}$, and the $t$ has the exact same meaning as the $x$ has in $f(x)$: it's the name of the independent variable of this function. We use $t$, not $x$, because $x$ is playing a role with the function $f$, and we don't want the same letter playing two different roles. $\endgroup$ Feb 15, 2012 at 6:19
  • $\begingroup$ @Luther: Remember: a function is just a rule that assigns to any (valid) input an output. While the definition of the function $\mathcal{F}$ may seem complicated, or unnatural, it is nonetheless a valid definition that gives you a function. This definition does not involve the notions of primitives, the FTC, or anything like that. Those notions come in later, after we prove that this function "happens to be" an antiderivative for $f$. In this respect, the $t$ of $\mathcal{F}$ is much like the $h$ in the definition of $f'(x)$; (cont) $\endgroup$ Feb 15, 2012 at 6:21

This started because I read somewhere that there is no such thing as an indefinite integral. An integral should be able to represent an area and if an indefinite integral has no bounds then it’s not an area.

Upon investigating further I found that those who were saying that there’s no such thing were then going on to show that the indefinite integral notation with no bounds at top and bottom of integral symbol is “shorthand” for the following formula, which they call the first part of the FTC and which is said to be the thing that connects the concept of differentiation/antidifferentiation to the concept of the integral.

A = F(x) = ∫f(t)dt[a, x]

I re-read Courant p. 110 http://ia700700.us.archive.org/34/items/DifferentialIntegralCalculusVolI/Courant-DifferentialIntegralCalculusVolI.pdf and had more insight into my biggest confusion, which is: what is t doing in there?

I believe the key to it is that the function notation has to be understood carefully. I still don’t understand the big picture and I’m still working at it.

The function in the above formula that is being integrated is y = f(t). My mind didn’t want to accept F(x) on the left with f(t) on the right. But then I remembered that the statement is being made that an antiderivative of f(t) is an integral, an area, as a function of the integral’s upper bound which is x. So my confusion is that y = f(t) is a function of t but F(x) which is both a primitive and an area is a function of x.

I think this is the answer I was looking for, because what stimulated the question is that it seemed people were saying that there is a way of seeing the primitive as an integral of something so this is it. And this is the FTC or the first part of it.

As to the drawing, it is easy to understand on its own terms but I now want to connect it to the above argument—so where’s the t?

Thanks to everybody who answered and commented so far including those who say I go on too long, that is true. Hopefully this is a better summary and includes a hint to where my answer might lie.

Here's the illustration from Courant & Robbins 1996 (he uses u instead of t): http://lutherspics.blogspot.com/

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    $\begingroup$ Don't let the t confuse you. When you write $\int_a^x f(t) dt$, then $a$ is fixed, and each value of x gives a value for the integral; $t$ is the variable of integration, taking values on the $x$-axis, between $a$ and x. We use another letter like t because inside the integral, x is already fixed (it is part of the bounds of the integral). If we changed t to u or anything else, nothing changes. But using x would be confusing because then we would have 2 x's that have different meanings (one is the upper bound of the integral, other is the variable of integration). $\endgroup$
    – Ted
    Feb 15, 2012 at 7:34
  • $\begingroup$ @Ted, "...each value of x gives a value for the integral..." because the area F is a function of x. "... inside the integral, x is already fixed..." Because if A = F(x) is fixed at some actual value, it's because x is fixed at some value and F is a function of x. I understand t could be u but not x. I still don't understand why it's called the variable of integration in this case. If this case is showing an antiderivative as a definite integral, then calling t by that name in this case is confusing. Seems like x is the real variable of integration here. $\endgroup$
    – Luther
    Feb 15, 2012 at 9:42
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    $\begingroup$ The variable of integration is the variable that takes values indicated by the bounds of the definite integral. In the definite integral $\int_1^2 x^2 dx$, $x$ is the variable of integration, taking values from 1 to 2. It's the same for $\int_a^x f(t) dt$: $t$ is the variable of integration. Both $t$ and $x$ denote points on the $x$-axis. (Just because it's called the $x$-axis doesn't mean we can't use other letters for its points.) Note that the value of a definite integral does not depend on the variable of integration: $\int_1^2 t^2 dt$ is the same as $\int_1^2 u^2 du$. $\endgroup$
    – Ted
    Feb 15, 2012 at 16:17
  • $\begingroup$ A link got left out of my question/answer: "As to the drawing, it is easy to understand on its own terms but I now want to connect it to the above argument—so where’s the t?" <math.stackexchange.com/a/15302/24956> $\endgroup$
    – Luther
    Feb 16, 2012 at 4:59
  • $\begingroup$ In the diagram you linked to, there is a curve labelled $f(x)$, and then also a point labelled $x$ on the horizontal axis. Those two $x$'s are NOT the same! The $x$ on the horizontal axis is a specific point on the axis, which we are interested in when defining the function $A(x)$. On the other hand, $f(x)$ denotes an entire curve and hence, in that context, there is no specific value of $x$. Because these uses of $x$ are different, we really should use two different letters. So the curve really should be labelled as $f(t)$ (or any other letter) instead of $f(x)$. Does that help? $\endgroup$
    – Ted
    Feb 16, 2012 at 5:52

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