Integrating left to right versus right to left.

Question

OK, I understand that when integration is done left to right with respect to x increasing left to right (dx is positive), that the answer is positive, and vice versa when integrating right to left. But, as a reality check, there is something that I have never understood (or, understanding, I find so counterintuitive as to want to question it): Interpreting the result as area, one and the same area is either positive or negative not depending on what it is, but on how it is counted. I recognize that it is a logical extension of, say, -1 versus +1 on the number line, but at least -1 and +1 occupy different positions! Whenever I've brought this up, I tend to get blank looks, like, "how could you be so stupid", or "huh", but, clearly it does not keep me from being bothered by it, so I'll ask here ... once again! To me, apart from what the mechanics of integration require, a more logical "positioning" of the negative area would be its reflection across the x-axis.

Answer 1

In short, the answer to your question is simply a matter of convention.

Suppose we have a rectangle of height $4$, whose base lies on the $x$-axis, beginning at $x=2$ and ending at $x=2.5$. If I asked you to find the area of this rectangle, you'd probably do the following computation:

$$4\cdot(2.5-2)=2\tag{$*$}$$

and you'd be correct. By convention, our number line increases as we move to the right, so to find the positive value of the base of the rectangle, we subtract its leftmost point from its rightmost point. If we changed the convention of our number line, we would also have to change our subtraction to obtain a positive result.

The Riemann integral is first defined only for intervals $[a,b]$ with $a<b$ (that is, with $a$ to the left of $b$). This definition is:

$$\int_a^bf(x)dx=\lim_{|P|\to 0}\sum_{i=0}^{n-1} f(x_i^*)(x_{i+1}-x_i)$$

where $|P|$ is the mesh of the partition $P=\{x_0,\ldots,x_n\}$, with $a=x_0<x_1<\ldots<x_n=b$, and $x_i<x_i^*<x_{i+1}$. The motivation for this definition, and not one in which we take $x_{i}-x_{i+1}$ in the sum is found precisely in the convention that caused us to calculate as we did in $(*)$. Then, having defined the Riemann integral for intervals $[a,b]$, we define the symbol $\int_{b}^af(x)dx$ to be:

$$\int_b^af(x)dx=-\int_a^bf(x)dx$$

Distributing the negative sign through the sum, you can think of this as making a definition

$$\int_b^af(x)dx=\lim_{|P|\to 0}\sum_{i=0}^{n-1} f(x_i^*)(x_{i}-x_{i+1})$$

It would be completely consistent to do calculus on a 'flipped' number line, where positive numbers are to the left of $0$ so that $a<b$ if and only if $a$ is to the right of $b$. Then we could integrate from right to left and obtain positive areas.

Answer 2

Although integrals can often be explained as "area", you are going wrong when you assume that integral is area. That is not true. Yes, they are related and numbers sometimes coincide.

But take a look at sine integral - it's zero, although the circumscribed areas are clearly non-zero. Does that seem O.K. to you? You have to accept that integral is not area. It can behave differently and it can have different signs.

Have you had physics in school? If you are familiar with potential energies, this might help.

Take a look at such situation. Object with mass $m$ is at height $a$. It is lifted up to a higher height $b$. Force of gravity is $F=mg$. The object-Earth system has gained additional potential energy: $$\Delta U_1 = \int\limits_a^b mg \mathrm{d}h = mg(b-a)$$

If the object was instead lowered from b to a then $$\Delta U_2 = \int\limits_b^a mg \mathrm{d}h = mg(a-b) = -\Delta U_1$$

Wither way the change in potential energy is the familiar $mg\Delta h$, but the change in height $\Delta h$ is opposite and so is the change in potential energy. Hope that's understandable and alright to you.

What I want to show, that integral is not the area function. It's more like accumulating operator. In the gravity case it accumulates energy by overcoming gravitational force while the argument $x$ (height) is changed. If you change it in other direction, the situation becomes obviously opposite.

Yes, if you take positive function and integrate it from left to right, you accumulate area and therefore you get the area. But if you go backwards, you are more like subtracting that area, therefore you get the minus sign.

Answer 3

I will try to explain it for the Riemann-Intgral: If you look at the definition of R-Integral you will notice that the term $(a-b)$ plays an important role. Namely $(a-b)/N$ gives you the length of the intervalls over wich the values of the function are avaluated and then summed. Here we are integrating over $[a,b]$. If we invert the direction of integration the previous expression changes to $(b-a)/N$ wich is equal to $-(a-b)/N$. Informally the Integral is given as the value taken in the interval $(a-b)/N$ times the Value that it takes in a precise point inside that interval. Hence if we invert $a$ and $b$ these values will be multiplied by $-(a-b)/N$ and hence the Integral with $a$ and $b$ inverted becomes negative