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

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.

Solution 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?

Solution 2:

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

Solution 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.