So I'm refreshing on integration and venture across a property of definite integrals I've been taking for granted for quite some time:
\begin{equation}\int_{a}^{b} f(x) \mathrm{d}x =-\int_{b}^{a} f(x) \mathrm{d}x \end{equation}
Stewart's Calculus claims this property is owed to $\Delta x= \dfrac{b-a}{n}$, where $\Delta x$ becomes negative when you flip $a$ and $b$. If this is true, then it would be true for the (I know it's not the exact) definition of a definite integral:
$$\int_{a}^{b} f(x) \mathrm{d}x =\lim_{n \to \infty} \sum_{i=1}^{n}f(x_i^*)\Delta x$$
So I just did a quick verification of a simple definite integral of $x^2$ from $0$ to $1$. Using the Riemann sum definition I got $1/3$ and switching the limits of integration I got $-1/3$ as expected. However, I noticed that if you take $\Delta x$ to always be positive, in the sense that it's a width of a rectangle and couldn't possibly be negative, then the answer either way would be $1/3$. Which, in conclusion, brings me back to my question: wouldn't it be simpler to take $\Delta x$ as a distance or width (always positive) and eliminate this property of integrals? Leaving us with:
$$\int_{a}^{b} f(x) \mathrm{d}x =\int_{b}^{a} f(x) \mathrm{d}x$$
To me this makes more sense, because the area under a curve should be maintained when calculated from any direction.
Best Answer
$\int_a^b f$ is a misleading notation. What is really defined is the integral over the interval, the set. There is no "orientation" involved. Think of this as $\int_{[a,b]} f$ instead.
Now, we define the symbols $\int_a^b f:=\int_{[a,b]} f$ and $\int_b^af:=- \int_{[a,b]} f$. Why? Because this is convenient. This makes a lot of formulas easier to write, and a lot of cases can be condensed in a single statement (for instance, as commented, the change of variables formula).