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 $\dfrac{1}{3}$.
 A: 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.
A: 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).
