## A matter of interpretation

In calculus 101, amongst the first things we learn, is that the derivative a function is the slope of the tangent to the function, that is, the instantaneous slope at some point on the function. We have, for some function $F$ that the derivative $f$ is given by:

$\displaystyle\frac{\partial\:F}{\partial\:x}=\lim_{\Delta\to{}0} \frac{F(x+\Delta)-F(x)}{(x+\Delta)-x}=\lim_{\Delta\to{}0}\frac{F(x+\Delta)-F(x)}{\Delta}=f$

So the formulation looks like a slope, and it is taught that it is a slope as well; all the concepts surrounding differentiation are expressed in terms of slopes of tangents, and that’s OK, because that’s what they are.

But suddenly, in calculus 201, we learn how to find the anti-derivative of a function, also known as the integral. But the metaphor changes completely: we’re know talking about the area under the curve. Wait. What?

Oh, so the inverse of the derivative, the slope, is not height, but area. Make sense. Or does it? If we look at the Riemann sum, it is usually noted as:

$\displaystyle \sum_{i=1}^n f(x_i) \Delta_{x_i}$

where the $x_i$ are positions where the function $f$ is evaluated, with a space of $\Delta_{x_i}$ before the next point, $x_{i+1}$. We can reparametrize the expression with a fixed variation $\Delta_x$:

$\displaystyle \sum_{i=1}^n f(a+i\Delta_{x}) \Delta_x$

$\displaystyle \sum_{i=1}^n f\left(a+i\frac{b-a}{n}\right) \frac{b-a}{n}$

and taking the limit:

$\displaystyle \lim_{n\to{}\infty}\sum_{i=1}^n f\left(a+i\frac{b-a}{n}\right) \frac{b-a}{n}=\int_{a}^{b}f(x)\:\partial{}\:x$

gives us, if the limit of the sum exists, the definite Riemann integral of $f$, the function $F$, on the interval $[a,b]$. This process is unfortunately taught as the limit of the sums of the areas under $f$, which makes no sense. Of course, you can interpret the product $f(x) \Delta_{x}$ as an area, a rectangle, and the (limit of the) sum of the rectangles as an area. While this is a valid interpretation, it’s not the most natural one.

If $f$ is the slope of $F$, wouldn’t it be more natural to interpret the integral as a height variation of $F$ on a given interval (for the definite integral) or the general expression of the height variation of $F$ for the indefinite case? In this interpretation, the Riemann sum becomes the limit of the sum of height variations; walking up and down $F$; an interpretation of the integral that does not involve the area, which bears no immediate relation to the slope (safe for the accidental form of the equation) but remains entirely in the conceptual space of slopes and heights. And I think that’s how it should be taught in schools.

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Interpreting the Riemann integral as the inverse of computing slopes, as computing the total height variation between two points (in the definite case) leads to an easier comprehension, or at least, better geometric intuition, when we consider more advanced integrals such as, say, the surface integral. In a surface integral, trying to understand the problem as sums and differences of volumes is not as helpful as understanding it in terms of gradients and curvature.

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Also, we can find the definite integral quite directly from the definition of the Riemann integral:

$\displaystyle \int_{a}^{b} f(x)\:\partial{}x=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i)\Delta_{x}$

$=\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} \frac{F(x+i\Delta_x)-F(x+(i-1)\Delta_x)}{\Delta_x}\Delta_{x}$

$=\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} F(x+i\Delta_x)-F(x+(i-1)\Delta_x)$

But this is a telescoping sum, where terms cancel:

$=\displaystyle \lim_{n\to\infty}F(x+n \Delta_x)-F(x)$

$=\displaystyle \lim_{n\to\infty}F(a+n \frac{b-a}{n})-F(a)$

$=F(b)-F(a)$

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Even if we forget about the area interpretation of the Riemann sum, we can derive the notions of “left” and “right” sum, or even middle point; suffice to reparametrize the sum. For:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^{n}\frac{F(x_{i+1})-F(x_i)}{x_{i+1}-x_i}$

To get the “left” sum, we put:

$\displaystyle x_{i+1}=a+i\frac{b-a}{n}$

$\displaystyle x_i=a+(i-1)\frac{b-a}{n}$

To get the “middle” sum, we have:

$\displaystyle x_{i+1}=a+\left(i+\frac{1}{2}\right)\frac{b-a}{n}$

$\displaystyle x_i=a+\left(i-\frac{1}{2}\right)\frac{b-a}{n}$

and, finally, the “right” sum:

$\displaystyle x_{i+1}=a+(i+1)\frac{b-a}{n}$

$\displaystyle x_i=a+i\frac{b-a}{n}$