Reversing the Bounds of a Definite Integral

Let us say we know that $\displaystyle\int_{a}^{b} f(x) \, \mathrm{d}x = C$ for some real $a,b$ and function $f(x)$. So what does this represent, graphically?

The area under the curve $f(x)$ in the interval $x=a$ to $x=b$, of course. What if we wrote $$\displaystyle\int_{b}^{a} f(x) \, \mathrm{d}x = C,$$ instead? That is, we just reversed the bounds?

Would it still be $C$?

No! In fact, by definition, it is $-C$. This is so that the two sides are reversed and the fundamental theorem of calculus works out nicely.

Some discussion of "backwards" integrals

We've seen how to define a definite integral $\int_a^b f(x)\,dx$ when $a\le b$ (so that $[a,b]$ is a nonempty interval), but there is also a convenient definition we can make when the endpoints are "backwards". The question we want to think about now is:

When $a > b$, how should we interpret the "backwards" definite integral $\int_a^b f(x)\,dx$?

One way to "derive" what the answer "should" be is using the additive interval property, $\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$, which we understand and know is true when $a\le b\le c$. The question is: what if $a,b,c$ are not in order? There's a chicken-and-egg sort of game here: we need to define backwards integrals to make any sense of this! But instead of grasping for straws, at least we can use the "to-be-determined generalized additive interval property" as a guide.

Yes! In general, we define $\int_a^b f(x)\,dx$ as $-\int_b^a f(x)\,dx$, when $a>b$. This definition allows you to generalize the additive interval property, $\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$, to allow $a,b,c$ to be any real numbers, not necessarily with $a\le b\le c$. All of this will be convenient when discussing integral functions and FTC later on.