# Changing bounds of definite integral

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)

• You don't. That's the convention for defining the integral when $a>b$. – Lord Shark the Unknown Dec 29 '18 at 11:19

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 $$\int_b^afdx=-\int_a^bfdx$$.
• Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $\varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $\varphi$. – Silent Dec 29 '18 at 11:34