Show $\int_a^b f(x)dx = - \int_b^a f(x)dx$. How to show $\int_a^b f(x)dx = - \int_b^a f(x)dx$ using the definition of Riemann integration?
$\int_a^b f(x)dx :=  \lim_{||P(x,t)|| \rightarrow 0} \sum_{i = 1}^{n-1} f(t_i) (x_i - x_{i-1}) $, where $P(x,t)$ is a tagged partition of $[a,b]$. But the definition of a partition assumes $ a = x_0 \leq x_1 ... \leq x_{n-1} \leq x_n = b 
 \Rightarrow a \leq b$, so then how is $\int_b^a f(x)dx$ defined if $ b > a$?
 A: For the usual definition to make sense, you need a non-empty interval $[a,b]$ (which implies $a<b$) and then you have a definition for the definite integral of $f$ over $[a,b]$, denoted as:
$$\int_a^bf$$
Notice that this definition does not make sense for $b=a$ (since there is no interval to partition) and not for $b<a$ (since the partitioning requires $a<b$; that's how intervals work).
Using this definition, you can prove the following important property:
$$\int_a^bf = \int_a^cf + \int_c^bf \tag{$*$}$$
With the usual definition, this property holds only for $a<c<b$. To allow more flexibility, one could want to give meaning to integrals of the form $\int_a^af$  and $\int_b^af$ (with $a<b$).
If we define them as:
$$\int_a^af := 0 \quad \mbox{and} \quad \int_b^a f:=-\int_a^b f$$
then the formula $(*)$ remains valid and holds in general, for all $a,b,c$.

How to show $\int_a^b f(x)dx = - \int_b^a f(x)dx$ using the definition of Riemann integration?

Long story short: this is usually a definition, not a property you prove. We define it in such a way that another important property remains valid in general.
