How do I convince my students that the choice of variable of integration is irrelevant?

Question

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.

Answer 1

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?

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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 :)

Answer 6

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.

Answer 7

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.

Answer 8

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

Answer 9

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.

Answer 10

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 $g(t)$ .

Therefore

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

Similarly

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

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

Answer 11

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.

Answer 12

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.

Answer 13

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.

Answer 14

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.

Answer 15

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

Answer 16

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?

Answer 17

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.

Answer 18

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.

Answer 19

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.

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