# Solving an integral when using a dummy variable

I'm trying to solve an integral and I don't remember how to when a variable is in the integral symbol, if that makes sense. I'm not sure what the correct terminology is.

So say I have an integral like:

$$f(x) = \int_{-3}^{x^{3}} t^2 dt$$

How do I solve it? What do I have to do differently?

• Were you only required to do this integral, or was this embedded in a problem of taking the derivative of an integral, something like$$\frac{d}{dx}\int_{-3}^{x^3}t^2\,dt\quad?$$Because computing such derivatives can be done without having to evaluate the integral first (by one part of the Fundamental Theorem of Calculus). Commented Jun 15, 2011 at 4:22
• @Arturo: I wonder the same thing... too bad we couldn't get a clarification. Commented Jul 13, 2011 at 14:36

When you are computing the definite integral $$\int_a^b f(t)dt$$ your usual procedure has been:

(i) Find an antiderivative $F(t)$ of $f(t)$. That means, find a function $F(t)$ such that $F'(t)=f(t)$.

(ii) Then your "answer" is $F(b)-F(a)$.

The procedure is exactly the same if $a$ and $b$ are not constants, but are functions of some variable $x$. So (ii) remains unchanged.

In your particular integral, we have $a=a(x)=x^3$ and $b=b(x)=-3$. The "lower" limit of integration is a constant, which you can think of as a constant function, and the upper limit is the function $x^3$.

Let's calculate. We have $f(t)=t^2$. So one antiderivative of $f(t)$ is $F(t)$, where $$F(t)=\frac{t^3}{3}$$

Now do the familiar substitution process. We get $$\int_{-3}^{x^3}t^2dt=F(x^3)-F(-3)=\frac{(x^3)^3 -(-3)^3}{3}$$

We might want to simplify this to $$\frac{x^9+27}{3}$$

Comments: The variable $t$ is called a "dummy variable" roughly because ultimately it plays no role in the answer. You would get exactly the same thing if you had $u$ as the variable, or $w$, instead of $t$.

In principle, you could also have used the letter $x$ as the dummy variable. But don't ever do it when one or both of the limits of integration involves $x$. Although it is technically not wrong, the chances of getting confused are too high.

• That's a really great explanation! But just on the last part - why is it not technically wrong (whilst being visually confusing) to use x as the dummy variable, as well as the symbol for the upper int. limit? I would have thought it would be the other way around - easier to understand what's happening, but actually technically very wrong? I'm only new to realising the limit variables and integrand variables should be/sometimes are written to be different. In my research on it, I've come to see that it's done because the two are in fact very different things - the x (in this case)... Commented Jul 14, 2020 at 18:23
• ...is an, as yet unchosen, single upper limit (could be 17, 3.75, 199, etc). The t though represents, in a way, 'each value' between the lower and upper limits, which all are to be plugged into the function (t^2 here), and multiplied by delta-x, and added, to form the limit of the Reinmann sum. That's of course a rough, over complicated explanation. But the point is that aren't the upper limit, and the dummy variable, two very different things - thus ifyou are being technically correct, one should in fact make sure the integrand variable (dummy) and upper limit variables are different? Thnx!:) Commented Jul 14, 2020 at 18:35

To be explicit, you can do this $$f(x) = \int_{-3}^{x^3} t^2\, dt = {t^3\over 3}\bigg|_{-3}^{x^3} = {x^9 + 27\over 3}.$$ There is nothing scary here; just treat $x$ as if it were a constant and you will be fine.

Easily done; let $u=x^3$, proceed with the integration as usual, and then replace the $u$ in your result with $x^3$.

• When I do so I get ((x^3)^2) - 9 as per the fundamental theorem of calculus. What am I doing wrong? Commented Jun 15, 2011 at 3:33
• Don't make that substitution. Commented Jun 15, 2011 at 4:41
• @In between: you are integrating wrong; the integral of $t^2$ is not $t^2$, it's $\frac{1}{3}t^3$ (plus a constant, but for definite integral evaluation you don't need the constant). Commented Jun 15, 2011 at 6:40