# Student forgets to remove dx after integrating

I am tutoring another US college student in a Calculus 1 class. Initially, she was having trouble with basic concepts, but after much prodding most of the conceptual difficulties seem to have been alleviated. However there is one persistent problem I've never encountered with another student: after integrating a function, she'll remove the $$\int$$ but leave $$dx$$ (or $$du$$ etc.). For example, she will do this:

$$\int 2x\, dx = x^2 + C\, dx$$

The $$dx$$ will hang around in her later calculations, and she'll express that something seems wrong (especially when none of the multiple-choice answers match - we're working with old exams), but I'll need to explicitly point out what happened before she fixes it. If I say "Why is the $$dx$$ still there?" she'll immediately realize that it was a mistake, though sometimes she will still ask why keeping it is wrong.

My first attempt at explaining the problem was "the $$dx$$ goes poof when the integral goes away" (this being how I learned it). My next was "the $$dx$$ marks the other end of what's being integrated, it's a boundary along with the $$\int$$". My third, after looking up what the $$dx$$ represented and finding What is $$dx$$ in integration?, was "it's like the $$\Delta x$$ in a Riemann sum, and when you evaluate the limit there is no $$\Delta x$$ still hanging around". All of these have been used more than once. Each attempt has stopped the forgetting for some more time but then she gets into the weeds of an integral and leaves the $$dx$$ on again while wrapped up thinking about something else.

How can I hammer this home so that it sticks? She is most confident in Riemann sums (that's part of why I used them in my third attempt) so an answer which leverages them would be appreciated but not required. We do online tutoring; anything that requires me being physically present will not work.

• Do you think there is an actual conceptual issue or just inattention to detail? The student is just asking “why is it not” but doesn’t seem to have any affirmative support justifying why it should stay. I think your second answer of it basically just being part of the operator is pretty much as simple as it gets here. Apr 21, 2022 at 23:30
• @Steve I'm honestly not sure. What concerned me is that she seems unable to grasp or even really repeat any of the explanation I give her; she opts to move on and then make the same mistake later. Apr 21, 2022 at 23:33
• Persistence. I kept pronouncing silent letters when I started learning french. I knew they were silent, but I couldn't catch myself because I couldn't think of everything at once. Cognitive overload. If you both keep at it, this problem will probably go away, but I doubt you need a more convincing argument. Apr 22, 2022 at 2:10
• If someone wants to give an experience-based answer saying that time will solve this, that might work! I've only been on-and-off tutoring calc since I took it myself (and mostly then for one-off students) so I don't have a great idea of what a typical long-term progression looks like. Apr 22, 2022 at 14:03

Since I teach physics, where units are very important, I would suggest considering $$x$$ to be a length (carrying units of meters).

The left-hand-side interpreted like an area has units of $$m^2$$.
The $$x^2$$ carries units of $$m^2$$... and $$C$$ carries units of $$m^2$$.
So, $$C dx$$ would carry units of $$m^3$$.

So, $$x^2 + C dx$$ would be inconsistent.

(A geometric construction to interpret the integral as an area may help support the units interpretation.)

(The student may need a reminder that the point of the constant of integration is that the derivative of that term should be zero. Having $$C dx$$ would be a strange construction for that purpose.

In other words, a "check" of the integration could be to "take the derivative of the right-hand-side".)

• I did end up going with this - especially the emphasis on making sure the +C was logically consistent, because that also addressed some other mistakes. The problem hasn't recurred and I'm crossing my fingers that it won't :) Apr 27, 2022 at 1:02

Give her repeated questions and give positive feedback, carrot, for doing it right. And negative feedback, stick, for doing it wrong.

There's no magic aha. She just needs attention to detail. And the above is how you train attention to detail.

Sometimes we may need to bend the rules a little bit. In a polynomial equation (the easiest), you could say that the $$dx$$ will turn into a $$x$$ on the other side of the equation, but in addition to integration rules.

Perhaps, at the initial class of integration, it would be useful to recall that a function $$f(x)$$ is actually a function of the form $$y=f(x)$$ and that it's derivative is $$dy/dx$$. So, when we integrate $$dy$$ will turn into $$y$$ and $$dx$$ will turn into $$x$$.

The easiest example could be $$dy/dx = 1$$, which turns out that $$dy = dx$$. So, when we integrate $$\int dy = \int dx$$ will be equal to $$y = x + c$$. An extra $$x$$ appears and the $$dx$$ disappears.

Another example would be $$y = a + bx + cx^2$$, where $$dy/dx = b + 2cx$$, thus $$dy = (b + 2cx)dx$$. So when we integrate, $$\int dy = \int (b + 2cx) dx$$ it turns into $$y = (b+cx)x$$, losing the integer because of "integration rules". Then, $$y = bx + cx^2 + Constant$$. An extra $$x$$ appears and the $$dx$$ disappears.

Of course this simple rule won't last too long, but at least a beginner student will master the basics and get use to take away $$dx$$. In addition, students should always remember that integrals are antiderivatives, and that trick helps a lot.

The way that I learned this, was quite visual, and involved graphs and talk of dimensions/directions so may not work if this is a totally new approach to thinking about integration:

When we intend to integrate

$$\int 2x\, dx$$

We need to make it clear what role the $$dx$$ is playing.

Look at the graph of the formula to be integrated: Here, we have to understand for ourselves that the formula shows how to get from an $$x$$ value to a $$y$$ value (i.e. by multiplying by two), but the formula $$y = 2x$$ itself says nothing about integration any enclosed area.

When we integrate, we are changing the dimensions we are interested in. Here it's from a one dimensional line, to a two dimensional area. Elsewhere it's more complicated, but let's focus on the easy for now.

Now this simple line is already splitting the graph into two areas, and when we integrate for most of it's just intuitive that we assume it's the area under the graph, e.g. towards the X-axis. But what in $$\int 2x\, dx$$ tells us to fill in that direction? It's the $$dx$$.

To reiterate, the $$dx$$ isn't decoration, it's direction. It's doesn't go poof and vanish. In this analogy, if our integration is a journey then the formula tells us the way we scale our step of X to the result of Y, but the integration shows us the direction of the steps. Each $$dx$$ steps along the X-axis in the X direction.

If might help to show the student this altered integration, and work through how to resolve it:

$$\int 2x\, dy$$

Here, we want to integrate with respect to $$y$$, not $$x$$. We're essentially flipping the graph around the gradient, and it points us to the fact that even though our gradient shows us how to get from $$x$$ -> $$y$$ (because we still have $$y = 2x$$) but our direction, and integration dimension has changed! If it merely vanished, we'd end up with the wrong result. This graph looks bonkers, but it shows what a $$dy$$ means in place of the $$dx$$. We're filling in a different direction.

Except there's a problem! We have nowhere in the formula for the $$dy$$ direction to point us; our scale and direction don't match!

To get to something we can integrate, we have to re-arrange our formula:

$$y = 2x \rightarrow x = \frac{y}{2}$$ $$\int \frac{y}{2}\, dy$$

Rearranging the formula lets us flip the axis of the graph back to what we're used to, but we're still keeping the integration direction of $$dy$$. We've just made the scale match for our convenience.

The fact they have to match shows you that use the $$dx$$ up. In polynomials, it directs all of it's matching 'scales' or powers to go up by one.

\begin{aligned} \int 1x^0+2x^1+5x^2\, dx &= x^{0+1}+2x^{1+1}+5x^{2+1} + C \\ &= x^{1}+2x^{2}+5x^{3} + C \end{aligned}

That $$+1$$ is from the $$dx$$ because that's the 'direction' the $$dx$$ takes us for polynomials. If you don't have a matching set of scales, you either need to rearrange to find one, or move everything except a $$1$$ out of the integration.

\begin{aligned} \int x+2x+5x^2\, dz &= (x+2x+5x^2)\times\int 1\, dz \\ &= (x+2x+5x^2)\times (z+C) \\ &= zx+2zx+5zx^2 + Cx+2Cx+5Cx^2 \end{aligned}

This concept of direction is powerful, as it applies to more than just polynomials. Look at the cycle of trigonometric integrations: Now we're armed with our 'direction' concept, we can apply it here given the direction in the above image:

$$\int \sin{x}\, dx = -\cos{x} + C$$

The $$dx$$ points us one formula around the circle, in the anti-clockwise direction (here anti-clockwise is pure convention - so long as we get between the right formulas it's ok). Taking that integration direction uses up the $$dx$$, in the same way adding one to the 'scale' of each polynomial term uses up the $$dx$$.

• I'm not sure I follow. A priori the indefinite integral is not about areas under a graph. but about antiderivatives. I think your explanation would make sense if we were talking about the definite integral. May 10, 2022 at 8:57
• @MichaelBächtold you are of course right, but I didn't think the concept of antiderivatives is a good entry to the topic of integrals. Especially as the OP says their student is aware of Riemann sums it looked like a good point to start from. It should be easy enough to go from their to antiderivatives if you extend the trig-circle I included May 10, 2022 at 11:25
• I'm not so sure the passage (in terms of notation) from the definite to indefinite integral is so trivial. It essentially boils down to the question of why we use almost the same notation as for the definite integral. (And sorry to be nitpicking, but that circle requires a multiplication with dx when applying $\int$.) May 10, 2022 at 11:33