Why is Area Determined by Integrals?

If you had any kind of math skepticism as calculus student in high school or freshman year (or maybe you currently are), you may have wondered why we calculate areas under curves using integrals. Why the hell is area the opposite of differentiation? The relationship does not seem any intuitive. This is a very good question! I have asked it to myself when I first learned about this fact as well since my teacher did not give an attempt to explain it. And even if you have never asked it to yourself, I believe this post may provide you with some interesting insight in case you are the tiny bit intrigued.

Note that, in this post, I will not go into any “analysis” arguments, especially that the question is about area, which is a geometric concept. This means that I will not go into the Fundamental Theorem of Calculus (FTC) which comes in two versions, and which mainly describes the “reverse” relation between derivatives and integrals. Hence, this post will be friendly to anyone with basic knowledge of calculus concepts (limits, differentiation, integration).

Without further ado, let’s dive right into it. Let’s assume we have a function f(x) defined on some interval [a,b] and we wish to find the area under the curve (C) of f on this interval. Let’s consider the area function A(x) defined on [a,b] to be the area under f from a to x. What we will do is to show that A^\prime\left(x\right)=f(x) which will basically show that A\left(x\right) is nothing but an integral of f(x).

As you are probably familiar with, the way to find the area is by summing over areas of rectangles (usually referred to as upper and lower Riemann sums) which approximate the area, then by the process of increasing these rectangles to get better approximations, if we take the limit of their number to infinity, we get an exact value for the area. However, I will follow a slightly different approach since we have formulated the problem in a way of showing that A^\prime\left(x\right)=f(x).

Example Picture using GeoGebra.

Let us divide the interval [a,b] as usual using rectangles each two of which are separated by a small distance \Delta x, and consider one of the rectangles which has a corner at x and another corner at x+\Delta x. Now notice that the area of this rectangle is nothing but f\left(x\right).\Delta x, which is approximately equal to the area under the curve from x to x+\Delta x; this area under the curve is equal to the area under the curve from a to x+\Delta x minus the area from a to x, i.e., A\left(x+\Delta x\right)-A(x). In other words,

(1)   \begin{equation*}f\left(x\right).\Delta x\approx A\left(x+\Delta x\right)-A(x) \end{equation*}


or,

(2)   \begin{equation*}f\left(x\right)\approx\frac{A\left(x+\Delta x\right)-A(x)}{\Delta x} \end{equation*}

Notice that, I said that these areas are approximately equal to each other due to the small excess of area above the curve which is part of the rectangle as in the figure (notice that it can be a deficit instead of an excess in other cases). The way to get a better approximation of f(x) is to decrease this excess, which can be done by decreasing \Delta x (i.e., an increase of the number of rectangles dividing [a,b]). As \Delta x approaches 0 (i.e., the number of rectangles approaches infinity), we get better and better approximations until we reach equality when we take the limit. In other words,

(3)   \begin{equation*} \lim_{\Delta x\rightarrow0} f(x)=\lim_{\Delta x\rightarrow0} \frac{A\left(x+\Delta x\right)-A(x)}{\Delta x} \end{equation*}


But, f(x) does not depend on \Delta x, so \lim_{\Delta x\rightarrow0}{f\left(x\right)}=f(x), which yields

(4)   \begin{equation*}f(x)=\lim_{\Delta x\rightarrow0} \frac{A\left(x+\Delta x\right)-A(x)}{\Delta x} \end{equation*}

And, here it comes! Recall that the derivative of a function g(x) at x_0 is defined as \lim_{\Delta x\rightarrow0} \frac{g\left(x_0+\Delta x\right)-g(x_0)}{\Delta x}. Hence, we can see that the right-hand side of the above equation is nothing but the definition of the derivative of A(x) at x. We finally get the result we are after,
f\left(x\right)=A'(x).

Although the establishment of this result may look long, it is nothing more than realizing how the area of the rectangle can be approximated by the difference of the certain areas under the curve, then taking the limit as \Delta x approaches 0.

Now, we have shown that A is an integral of f, but an integral of a function also depends on the constant of integration. However, we know that the area is fixed and should not depend on any constant, so we need to find this constant of integration. To do this, we write A\left(x\right)=\int_{s}^{x}f\left(t\right)dt where s is to be determined; this form takes into consideration the constant of integration since we have boundaries for the integral. As usual, to determine the constant of integration, we need a condition. Notice that the area under (C) from a to a is zero, i.e., A\left(a\right)=\int_{s}^{a}f\left(t\right)dt=0. The only way we are guaranteed this integral to be 0 is to take s=a, because otherwise, we may never get zero (if the function is always positive for example, the integral will always be positive and only 0 at a).
Finally, we reach that
A\left(x\right)=\int_{a}^{x}f\left(t\right)dt

And so, the area under the curve from a to b, which is nothing but A(b) is \int_{a}^{b}f\left(x\right)dx (here we put back x in place of t since it is a dummy variable and we no longer have an x on the bounds of the integral).

Short question: Can you figure out why integrals give us the negative of the area if the curve is under the x-axis?

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