# area under a curve and units

If we introduce a unit of length like meter for $x$ and integrate the function $f(x)=x^2$ from $0$ to $2m$ we get $\dfrac{8}{3} m^3$. How can this be interpreted geometrically?

My initial thought was that $x^2$ gives the area of a square with edge length $x$, so maybe the area under the curve gives the volume of a cube with a base area equal to this square. But this is off by a factor of $\dfrac{1}{3}$.

• i think might be a pyramid already the base varies in size with height, but maybe I'm wrong.
– cand
Dec 15 '14 at 0:58
• Yeah, that's really curious. Is there a pyramid hidden in my problem somehow or is it a coincidence that the integral gives the volume of a pyramid?
– Marc
Dec 15 '14 at 1:23
• taking x a height, the area of the basis at height x is $x^2$, the integral is sum over all infinitesimal pieces and when all pieces are joited you get a pyramid with lenght 2m and height 2m, with make any sence that the formula give the volume of a pyramid, but same as before, i can be wrong.
– cand
Dec 15 '14 at 1:32
– Marc
Dec 15 '14 at 1:38
• i thinking something like Deepak as saying, something like cuting the pyramid into small enought cubes, the lenght of these small cubes dependes of the height on x, and then sumation of all these small "cubes" gives the volume of the pyramid :)
– cand
Dec 15 '14 at 2:02

When you do the integral you are doing $\int y \ dx$, so the units of $y$ should be meters as well. One way to think of it is that $f(x)=ax^2$, where $a$ has units of $m^{-1}$. This makes $f(x)$ have units of $m$ as it should. It just turns out that when measuring in meters the numerical value of $a$ is $1$. If you change to measuring in $cm$, you now have $\int_0^{200} \frac 1{100}x^2\ dx=\left. \frac 1{300}x^3\right|_0^{200}=\frac 83\cdot 10,000\ cm^2$ as it should. As you can see, you need to update the value of $a$ because it is carrying a unit as well.

• Thanks, but it seems a bit like a workaround. There was an earlier comment which has been deleted which suggested using a double integral. Maybe this is more natural. I don't know how to write it down yet though.
– Marc
Dec 15 '14 at 1:18
• @Marc In my deleted comment I was just saying that an expression with units of $m^2$ is already an area and thus could be considered an integral by itself. I couldn't think (and didn't want to think too hard about it) what kind of picture this implies.
– user137731
Dec 15 '14 at 1:25
• The point is that you shouldn't consider $m^2$ to be an area. It isn't, it is a length parallel to the $y$ axis. If you want to assign units, you need to do so in a consistent way. The answer comes out proportional to $m^3$, but that doesn't say it is a volume. Dec 15 '14 at 1:37
• Yeah, I see that this is a consistent way of looking at this. I am just wondering if (a) we get a more natural picture if we use a double integral and (b) we somehow get a geometrical interpretation to work which involves the volume. It is a curiosity that the calculation yields the volume of a pyramid.
– Marc
Dec 15 '14 at 1:42

It all depends on your interpretation (setup).

If your goal is to compute the $2$-dimensional area under the curve $y = x^2$ then $y$ must also have the unit of metre ($m$), which gives an area in terms of square-metre ($m^2$), as you would expect.

If, however, you are aiming for a "natural" 3-dimensional geometric interpretation, then you can think of it like so: The $y$-value gives the enclosed area of a square of side $x$. So the integral you computed actually gives you the enclosed volume of a square pyramid. If you take cross-sections across such a shape, you will get squares of different sides ($x$) and different enclosed areas $(y = x^2)$. Naturally, the calculated volume will have units of cubic-metre $(m^3)$. If you start the $x$ value off at something other than zero, you will be computing the volume of a square-pyramidal frustum (truncated square pyramid).