# Integral as “Area Under the Curve”

I'm a high school math teacher teaching an introductory calculus course, and I'm having a problem teaching one particular student about the physical definition of an integral.

The intuition is that it's the "area under the curve," and all but one of my students accept that this implicitly means "area under the curve down to the x-axis," but one student is hung up on thinking that "area under the curve" extends all the way down to $y = -\infty$.

I tried giving him the following proof:

Let $\int_0^1 f(x) \ dx$ be the area under $f(x)$ extending to $y = -\infty$. It is clear visually that $\int_0^1 f(x)-g(x) \ dx$ is the area between the functions $f(x)$ and $g(x)$. Now let $g(x) = 0$. We then have that $\int_0^1 f(x)-0 \ dx = \int_0^1 f(x) \ dx$ is the area under $f(x)$ extending to $y = 0$, contradicting our initial assumption.

He seems unconvinced by the above procedure. Is there any alternate phrasing I can use to convince him of this? I don't want him thinking that every integral evaluates to $\infty$ just because he's hung up on the wording.

I'm new to teaching and I've never had this misconception come up before, so I'm trying to fumble around with ways to properly explain this such that my other students don't doubt their correct intuition.

• I didn't downvote but I'm not sure I understand. Why don't you look at trapezium approximations or mid ordinate rule take it from there? – Karl Apr 19 '16 at 16:40
• Offer him a very simple example, $f(x) = 1$. Assuming he accepts that the integral of this function over $[0,1]$ is $1$, then if the integral represents the area of a region under the curve, then the only lower boundary that results in unit area is the $x$-axis. – Bungo Apr 19 '16 at 16:43
• Why not always say (for non-negative $f(x)$) "the area under the curve and above the $x$-axis" instead of the apparently problematic "area under the curve"? – André Nicolas Apr 19 '16 at 16:43
• The "area under the curve" means the area under the curve with the implicit assumption above the $x$-axis. There is no need to prove anything. In your case, I think you should seriously consider whether your student is really stuck on that idea or want to play a word game against you. – achille hui Apr 19 '16 at 16:44

If you remind them the integration is somewhat like a 'continuous' analogue to a 'discrete' summation, it would become quite obvious the integral of a zero function should be zero, similar to a sum of zero terms sequence. And that the zero value of the integral corresponds to 'no area' of a degenerate figure 'between' the $f(x)=0$ graph and the $X$ axis.

Show some physical applications, like integrating (one-dimensional) density into a mass (say a mass of a wire with varying diameter) or integrating a velocity over time into a distance travelled. They should make a clear need of additivity (twice the density - twice the mass, 10 mph more - ten miles farther in an hour).

In the next step show the sum of negative terms and the integral of negative function. And then a sum of an alternating-sign sequence (possibly reducing to zero) and an integral of an alternating-sign function (like sine). You can show a 3D example of earthworks and excavations (as an illustration, not calculation) with balancing volumes integrated from the heights and depths relative to ground level.

Then show that by additivity for a finite integral there's always a constant term which added to an integrated function makes the integral zero, and that corresponds to shifting the graph vertically so that 'positive' and 'negative' areas above and below axis make balance.

That should let them grasp an idea (along with formal definitions) how additivity works and where the zero 'should be' in integration – and especially why zero is at the axis level.

Regarding your proof, to be honest, I think the student is right to be sceptical as there are a couple of issues with it! Firstly your suggestion that it is 'clear visually' that $\int^1_0 f(x) dx -\int^1_0 g(x) dx$ represents the area between the curves may very well be the case, but unfortunately when it comes to doing arithmetic with infinite quantities what is clear visually is often not actually the case! Secondly, I assume that the expression I've written above was what you were really thinking of when you made the statement, but by writing it as $\int^1_0 f(x) - g(x) dx$, you've assumed that this new definition of integral distributes over addition (or subtraction), which is wrong — in fact, the contradiction you reach arguably works better as a refutation of this latter assumption than what you were originally aiming to do.

Really, I don't think you should be trying to 'disprove' the student's understanding of 'area under the curve' at all, as to an extent they're making a valid point: the points below the x-axis are 'under the curve'. The main problem with using that to make a definition is that it would always be infinite and therefore of limited use or interest. Also, the fact that everyone else is taking the integral as 'the area under the curve but above the x-axis' means that to do anything else would be doomed to cause major confusion.

If I were you, when he questions the meaning of area under the curve, I would simply admit that the phrase is a bit ambiguous and clarify that you mean only down to the axis.