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I will be TA this semester for the second course on Calculus, which contains the definite integral.

I have thought this since the time I took this course, so how do I convince my students that for a definite integral

$$\int_a^b f(x)\ dx=\int_a^b f(z)\ dz=\int_a^b f(☺)\ d☺$$

i.e. The choice of variable of integration is irrelevant?

I still do not have an answer to this question, so I would really hope someone would guide me along, or share your thoughts. (through comments of course)

NEW EDIT: I've found a relevant example from before, that will probably confuse most new students. And also give new insights to this question.

Example: If $f$ is continuous, prove that

$$\int_0^{\pi/2}f(\cos x)\ dx = \int_0^{\pi/2}f(\sin x)\ dx$$

And so I start proving...

Note that $\cos x=\sin(\frac{\pi}{2} -x)$ and that $f$ is continuous, the integral is well-defined and

$$\int_0^{\pi/2}f(\cos x)\ dx=\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx $$

Applying the substitution $u=\frac{\pi}{2} -x$, we obtain $dx =-du$ and hence

$$\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx=-\int_{\pi/2}^{0}f(\sin u)\ du=\int_0^{\pi/2}f(\sin u)\ du\color{red}{=\int_0^{\pi/2}f(\sin x)\ dx}$$

Where the red part is the replacement of the dummy variable. So now, students, or even some of my peers will ask: $u$ is now dependent on $x$, what now? Why is the replacement still valid?

For me, I guess I will still answer according to the best answer here (by Harald), but I would love to hear more comments about this.

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I don't get why this is different from understanding that the functions $f(x) = x^2$ and $f(z)=z^2$ are the same. Why involving integration at all? – Daniel R Jan 21 '14 at 14:48
My chain of thought: Before you start doing integration, you probably will need to understand what a function is. And since the symbol chosen as the free variable in a function doesn't change the function, it all boils down to explaining this before even involving integration. I'm trying to understand what you mean here, don't be hostile please. – Daniel R Jan 21 '14 at 14:59
What's in a name? That which we call a rose By any other name would smell as sweet. – Edoardo Lanari Jan 21 '14 at 15:02
@Lano Is that from Math163: Calculus for English Literature Majors? – Superbest Jan 21 '14 at 19:02
@Lano unless you have synesthesia? – Michael Jan 21 '14 at 19:57

19 Answers 19

up vote 148 down vote accepted

Draw a graph of the function on the blackboard, showing $a$ and $b$ and a crosshatched area representing the integral. Put an $x$ on the horizontal axis. Erase the $x$ and put a $z$ there. Does that change the area? Erase the $z$ and put a smiley face there. Does the area change? Why/why not?

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Nice way of doing so! (+1) – ireallydonknow Jan 21 '14 at 14:37
This is a cute way, but don't you think it can be explained using logic rather than pictures? I have the feeling that when a teacher stands in front of the blackboard explaining things this way, he's relying on his authority as a teacher too much to convince his students. – Egbert Jan 21 '14 at 15:05
@Egbert The logical explanation would be exactly the same: No object changes just because we change the name by which we refer to it. It complicates the discussion a bit here that “the object” has a rather nebulous existence – I mean, what is an integration variable after all? – but naming invariance is the gist of the matter. I know people who have changed their name. But the person is still the same, to the extent that anybody remains the same. – Harald Hanche-Olsen Jan 21 '14 at 15:18
On the contrary, I would argue that this is using logic. Not all individuals have the ability to grasp logic out of thin air, providing pictures like this as a point of reference allows them to assign logic to something tangible. If I didn't understand math, I would have to take it on the teacher's authority that 1+1=2. However, if I saw a picture of 2 single apples being put together to form a group of 2 apples, I may be able to logically put together that 1+1=2. – Dave Jan 21 '14 at 16:25
This is a great answer, and the opposition suggested in some of the comments between pictures and "logic" seems specious. It is especially good because in most freshman calculus courses one of the key pedagogical points is that $\int_a^b f(x)dx$ is not defined to be $F(b)-F(a)$ but rather the signed area under the curve (the latter being made more precise using exhaustion/Riemann sums when needed). The signed area under the curve cannot possibly depend on the variable name. Students who think it does need reinforcement on what the definite integral means. – Pete L. Clark Jan 22 '14 at 4:24

Start by showing $$ \sum_{k=1}^{5}a_k = a_1+a_2+a_3+a_4+a_5=\sum_{j=1}^{5}a_j$$

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I really like this – Cruncher Jan 21 '14 at 18:48
That, in my opinion, is a very good example of it. Not only it is simple but also relevant to integral since both are summations, only one is for continuous and the other is for discrete. I remember how amazed I was when I realized this distinction. – Varaquilex Jan 21 '14 at 18:56
@Varaquilex, right, the good old counting measure! – Carsten S Jan 22 '14 at 15:02
Yes, I was going to post something like this. +1 for you. I think an even better example would be where the summand is actually an explicit function of the dummy, such as $\sum_{k=1}^{3}1/k = 1/1 + 1/2 + 1/3 = \sum_{n=1}^{3}1/n$ – MPW Jan 22 '14 at 15:07

I can tell you exactly why your students are confused. It is because when they are taught the indefinite integral, the variable inside the integral sign appears to be the same variable as the one in the result. But in fact, the indefinite integral is a shorthand- and the variable in the result logically appears in the limits of integration of the integral, not in the dummy variable used to integrate over. Their calculus professor probably skimmed over that, I know mine did.

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This confused me briefly - it took me a minute or so to realize that I was looking at a definite integral, not an indefinite integral. – Brilliand Jan 21 '14 at 20:46
oh, this answer is good and gets right to the heart (one of the hearts) of the problem. Also good for elucidating CDFs like $F(x) = \int \limits^{x}_0 f(z) dz$ – Trevor Alexander Jan 21 '14 at 22:38

I have had students with several years of calculus under their belt get confused about this, and I have found that a rather low-brow "explanation" is most effective: first, write down $$\int_0^1 t^2\, dt$$ on the board and have them calculate it. Then write down $$\int_0^1 z^2\, dz$$ and have them calculate it. This usually causes them to have an epiphany, and while they might not be able to articulate it correctly they are having the "right" epiphany; they realize intuitively that the important thing about a function is not its formula but the relationship between inputs and outputs that it asserts.

I would caution against some of the more verbose metaphors and analogies that others have recommended, and I would especially caution against explaining this by referring to the foundations (there is a reason why the very notion of a function emerged almost a century after calculus was discovered). Students (maybe all of us?) think in terms of examples, and the closer the example to their point of confusion the better.

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+1 for the advice to keep things concrete. This is probably the number one way I could improve my own teaching if I fully internalized it. – Pete L. Clark Jan 22 '14 at 4:39

In addition to the otherwise excellent ansaers so long:

How do you define the definite integral? By Riemann sums, probably. Then show the Riemann sum do not depend on the summation variable.

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I would have simply said, "Look at the answer. Do you see any x [or z or smiley] there? Ok, then how could the choice of symbol matter?" – Carl Witthoft Jan 21 '14 at 19:01
This answer is mathematically correct but looks a little shaky pedagogically (at least for the American freshman calculus classes I have met). The "Riemann sum" is a complicated object beyond the full comprehension of all but the very best students. Explaining things in terms of Riemann sums (rather than areas as in Harald's answer) is going to make things obscure to many students. Historically, you are using a concept from the 1800's to explain something that was well understood in the 1600's. – Pete L. Clark Jan 22 '14 at 4:33
Also note that the Riemann sum has lots of complicated $x$'s in it -- often some $x_i$'s and $\Delta x$'s and even $x_i^*$'s. Replacing all of these with $y_i$'s does not make it immediately clear on a syntactic level that you get the same thing: you have to understand what everything means. – Pete L. Clark Jan 22 '14 at 4:35

Well, I usually explain it without any mathematics.

Imagine a doctors office. Describe what happens, when somebody goes there assuming to have the flu.

The description doesn't depend on the name. The doctor may use the word patient or client, or something else, just as she wants. The complete process is in the office, the way to refer to the person stays in the office.

In that way, the integral is a closed thing like a doctors office, and in it we can use any name we want.

PS.: If your students are afraid of doctors, you may use another office :)

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Why is doctor "she" and not "he"? – Kyslik Jan 22 '14 at 7:04
@Kyslik it is possible for women to be doctors. – jwg Jan 22 '14 at 9:32
@jwg I know I hope I didn't offend anyone, I was just curious. In my native language "r" sounds "hard" or "strong" (we differ gender of objects by endigs) similar to german [der, die, das] but without (mandatory) article. Therefore I am usually confused how people differ gender in english language. I think @Falko should use "doctor" instead of "she", because in PS @Falko states doctors does it mean all doctors are women? (of course not, I am going in to too much detail). I like the answer itself. – Kyslik Jan 22 '14 at 11:38
Often in English when using an example of an abstract person we'll use a gendered pronoun for flavor. To continue referring to the person as "doctor" is taxing, and not necessary. Frequently, you will see he/she, but she is more progressive as for long in our history we used he and him. They is an acceptable neutral pronoun but it has the problem of also being used to refer to a group of people. – VoronoiPotato Jan 22 '14 at 14:17
Personally, I don't think this would be such a good explanation. Some students may not understand the relation between a doctor and an integral (I'm serious here, no joking) because they just can't handle that level of abstraction. There are students having difficulties only to understand the notation $$\sum_{k=1}^3a_k.$$ If they hardly understand the relation between this notation and a sum, how would they ever understand the relation between a doctor and an indefinite integral? This may be a good explanation to make it more clear for people who already understand the issue, not for others. – barto Jan 23 '14 at 14:43

From a programmer's point of view $\int_a^b f(x)\,dx$ means the "function" which takes two numbers ($a$ and $b$) and other function $f$ as arguments and return a number. $x$ is only the local variable, the scope of it is within the function of integration, it is not visible in global scope (outside the function of integration). So it is not important how it is named, the "program" will work anyway.

Here is an example which illustrates these words (C-like syntax is used but this is not a real code, of course):

real_number integral(real_number a, real_number b, real_number f(real_number))
    // function scope, local
    // x is visible only here, so its name doesn't matter
    real_number result;
    real_number x;
    real_number dx;

    result = 0;
    x = a;
    dx = (b - a) / infinity;
    while (x < b)
        result += f(x) * dx;
        x += dx;

    return result;

// global scope
// x is already not visible here

I = integral(a, b, f);

Possibly this explanation may be useful.

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You'd be surprised how many programmers don't understand the concept of a function as a first class member of a programming language. – Chuu Jan 21 '14 at 22:28
I've never heard the phrase "first class member" but I think I know what a function is. – josh314 Jan 21 '14 at 22:32
@Chuu Then one can explain them this concept using the concept of integral. The concept of recursion is obtained as a useful side effect in this case. – Constructor Jan 21 '14 at 22:39
@josh314 – Chuu Jan 21 '14 at 22:51
I just knew that someone would write a program to show this :P – Navin Jan 23 '14 at 5:03

Discuss $$ \int_a^b f $$ in class. Why it is sometimes a useful definition, that $f$ is the name of a function, compare it to the $$ \int_a^b f(\mathcal{V}) d\mathcal{V} $$ and give them examples $$ \int_a^b 1 = b-a \qquad \int_a^b \cos = \sin b- \sin a \qquad \int_a^b \operatorname{id} = \frac{b^2-a^2}{2} $$ The main problem is that sudents think that (expect, are used to) cosine should be called $\cos(x)$

Never say that $f(x)=\sin(x)$ is a function, $f=\sin$ is the function, $f(x)=\sin(x)$ is what comes out if we plug a fixed $x$ and so on

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For that matter, how do you explain to them that x= 5+3 is the same as z=5+3 ? If they can't understand what a variable is, you're kinda stuck.

This example as well as the original question have numeric solutions, not algebraic. That's what they need to comprehend.

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But x=5+3 is not the same as z=5+3! – jwg Jan 22 '14 at 9:33
But $\{ x\in\mathbb R\ |\ x=5+3 \}$ is the same as $\{ z\in\mathbb R\ |\ z=5+3\}$, now go and explain set theory and the axiom schema of specification to your students! – Christoph Jan 22 '14 at 11:32
@jwg Since, as I said, this is about finding an answer, not about the equation itself, the answer is the same regardless of what you name the unknown. – Carl Witthoft Jan 22 '14 at 12:44
@CarlWitthoft, if you can't understand what a variable is, you're kinda stuck. – jwg Jan 22 '14 at 12:47

I think for a mathematical proof this should work :

Take a function $f(x)$ and let us assume it's indefinite integral is $g(x)$ . So if we assume a function $f(t)$ , its indefinite integral should be $g(t)$ .


$$\int_{a}^bf(x)dx = [g(x)]_{a}^{b} = g(b) - g(a)$$


$$\int_{a}^bf(t)dt = [g(t)]_{a}^{b} = g(b) - g(a)$$

It doesn't matter which variable we are using but finally we are going to use the value of $a$ and $b$ .

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This can be looked upon in two ways:

  1. In your case, it seems like $x$, $z$ and $☺$ are in fact different names for what can perhaps be considered to be the same variable. If that is the case, you could maybe explain that changing the name of a variable has no impact on the expressions it is involved in. It doesn't matter how a variable has been intended to be used (here, the intended use is likely to depend on the variable name), it only matters how it is really used, and here, $x$, $z$ and $☺$ are used in the same way. If we besides can consider $x$ = $z$ = $☺$ to be true, these are truly interchangeable variables (or variable names?), and the identity you want to show should follow naturally.

  2. On the other hand, if $x$, $z$ and $☺$ are different variables that in some way depend on each other, and $f$ is evaluated differently depending on if it is given $x$, $z$ or $☺$ as argument (i.e., it is possible to write $z$ and $x$ as functions of each other: $z = z(x)$ and $x = x(z)$, and $f(x) = f(z) = f(z(x)) = f(x(z))$), we then have

    $$\int_{\displaystyle x_{\text{min}}}^{\displaystyle x_{\text{max}}} f(x)\,\operatorname{d}x \,=\, \int_{\displaystyle x_{\text{min}}}^{\displaystyle x_{\text{max}}} f(z(x))\,\operatorname{d}x$$ $$=\, \int_{\displaystyle z(x_{\text{min}})}^{\displaystyle z(x_{\text{max}})} f(z)\,\frac{\operatorname{d}x}{\operatorname{d}z}\operatorname{d}z \,=\, \int_{\displaystyle z_{\text{min}}}^{\displaystyle z_{\text{max}}} f(z)\,x'(z)\operatorname{d}z$$

    and we see that $\int_a^b f(x)\,\operatorname{d}x$ and $\int_a^b f(z)\,\operatorname{d}z$ are not equal in the general case.

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I believe the problem lies in the fact that the symbol $x$ is often used as the irrelevant variable when defining a function:

$f : {\mathcal D} \rightarrow {\mathcal C} , x \mapsto x^2$

First I would make sure that students understand that the above and any of the following are equivalent:

$g : {\mathcal D} \rightarrow {\mathcal C} , a \mapsto a^2$

$h : {\mathcal D} \rightarrow {\mathcal C} , b \mapsto b^2$

in that they define the same function (same domain, codomain, mapping) albeit with different names f, g, h.

Secondly, once they fully grasp the nature of:

$f : {\mathcal D} \rightarrow {\mathcal C} , x \mapsto x^2$

I feel like they should understand that, by analogy/simmetry, also:

$\int_l^u f(y) dy$ and $\int_l^u f(z) dz$

are equivalent. Note that I defined the function using the symbol $x$ and then I used other symbols ($y, z$) as the integration variables.

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I am TA myself and I often have these issues of explaining mathematical concepts into easy to understand concepts as most of my students are first year uni students and the subject is compulsary so most of them don't even want to be there.

My advice is to use concepts that they can relate to.

I think the problem students find is when you start using letters of the alphabet they panic because it becomes abstract. Instead of using abstraction use metrics they understand such as metres, kilograms, etc. Give examples of how integration can be used using these metrics.

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The most straightforward way to understand why changing the symbol for the variable of integration doesn't change the answer is to unwrap what the mathematical expression for a definite integral means.

The (Riemann) integral of a function f from a to b is defined as a type of limit of Riemann sums. If you write the expression for this limit as $$\int_a^b f(z)\ dz$$ instead of $$\int_a^b f(x)\ dx$$, does this change the limiting value? Of course not.

Indeed, notice that if you say it out in words, "The Riemnn integral of f from a to b" doesn't even use 'x'. The only reason it would appear there is if you are accustomed to calling functions f(x) instead of f, but then it should hopefully be clear that changing the name of the independent variable from x to z will not change the function and hence would not change the integral.

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$x$ eventually gets replaced by $a$ and $b$.

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Maybe it helps to investigate the wording: integration variable is just a fancy name for what we used to call placeholder in elementary school when we solved

 3 + _ = 5

and used an underscore or an empty box as the placeholder. Isn't it obvious then that the symbol (or variable name) cannot have an effect on the solution?

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I think something to think of here is the fact that in teaching calculus (especially the definite integral; which from my experience is a second semester calculus class) the students should already be aware of the fact that x is a variable and the choice of letter (or symbol) is completely arbitrary. I would be more worried about the students that can't see this and would see it as them being in the wrong class. If they don't understand the fundamental basis of a variable they probably shouldn't be in a calculus class.

I would venture to say that this shouldn't be/isn't a problem for a university level calculus class.

If you truly are worried about this, I would suggest not focusing on the topic but instead showing it through examples; continually substitute in whimsical pictographs in place of the constant x/y/z. It would be entertaining to the students and would drive the point home without you actually having to focus on it.

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Even if they shouldn't be there, there they are. Can't just dismiss them with "you should know this" and move along. – Mark Fantini Jan 21 '14 at 19:57
I understand. I hope the class goes well for your sake. It seems like your going to be getting LOTS of questions if they can't handle an arbitrary variable. God help them when they hit infinite series. I should say that my professor did "move along" with MUCH heavier material. They must have it pretty easy. ;P – user122845 Jan 22 '14 at 5:35
My teachers did also move along with much heavier material as well. It does not mean it should be done. – Mark Fantini Jan 22 '14 at 17:33

Show examples with definite integrals! And give another example of a function:

say f(x) = sin(x). The function is the same (i.e. representing the same thing) if we replace x with y, i.e. f(y) = sin(y). This gives them an example that they should be familiar with and puts it into context that what symbol you use for a variable doesn't matter.

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I have often thought about using this approach but never dared to actually try it:

Every time you perform manipulations on the blackboard using a dummy variable, change the symbol used for the dummy in each instance of the summation or integral.

$$ \int_{x = \frac{\pi}{4}}^{x = \frac{\pi}{2}} 2 \csc x dx = 2 \int_{\phi = \frac{\pi}{4}}^{\phi = \frac{\pi}{2}} \csc \phi d\phi = 2 \int_{a = \frac{\pi}{4}}^{a = \frac{\pi}{2}} \frac{da}{\sin a} = 2\int_{t = \sqrt{2} - 1}^{t = 1} \frac{2t}{1 + t^2} \cdot \frac{2dt}{1 + t^2} = 2\int_{W = \sqrt{2} - 1}^{W = 1} \frac{dW}{W} $$

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This a very sure way to annoy your readers. – TZakrevskiy Jan 22 '14 at 23:10
Note that the last two steps should change the variable as they refer to different values. It does actually aid the readability if the first three steps did not change it, except to prove the point of the OP. – Mark Hurd Jan 24 '14 at 0:54

protected by Michael Greinecker Jan 22 '14 at 1:18

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