Changing bounds of definite integral

Question

I am using this baby Rudin definition for Riemann integral:

Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, \ldots, x_n$, where $$ a = x_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq x_n = b.$$ We write $$ \Delta x_i = x_i - x_{i-1} \qquad (i = 1, \ldots, n). $$ Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put $$ \begin{align} M_i &= \sup f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ m_i &= \inf f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ U(P, f) &= \sum_{i=1}^n M_i \Delta x_i, \\ L(P, f) &= \sum_{i=1}^n m_i \Delta x_i, \end{align} $$ and finally $$ \begin{align} \tag{1} \overline{\int_a^b} f dx &= \inf U(P, f), \\ \tag{2} \underline{\int_a^b} f dx &= \sup L(P, f),\\\, \end{align} $$ where the $\inf$ and the $\sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.

If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$, $$ \tag{3} \int_a^b f dx. $$

How do we get property $\displaystyle\int_a^b f dx=-\displaystyle\int_b^a f dx$ from this definition?

(Code borrowed from here)

Answer 1

You can't, because the definition you quoted deals with integrals for $a>b$. You have to further define that $\int_a^afdx=0$ and $\int_b^afdx=-\int_a^bfdx$.

Why you ask? Because it's logical. Negative sign is often asigned to changing the orientation. Also, it plays nicely with the "area" interpretation of integrals.