This certainly isn't a proof. But it does seem to make sense to me. You need to look at it as a line integral.
To compute $\int_0^3 f(x) dx$, we can start by choosing breakpoints
$\{0, 1, 2, 3\}$ for the interval $[0, 3].$
It's important to notice that the sequence goes from $0$ to $3$.
Then there exists
$\{\bar x_0, \bar x_1, \bar x_2\}$
such that
$$\int_0^3 f(x) dx = f(\bar x_0)(1-0) + f(\bar x_1)(2-1) + f(\bar x_2)(3-2)$$
To compute $\int_3^0 f(x) dx$, we start by choosing breakpoints
$\{3, 2, 1, 0\}$ for the interval $[0, 3],$
where the sequence now goes from $3$ down to $0$.
Then there exists
$\{\bar x_0, \bar x_1, \bar x_2\}$
such that
$$\int_3^0 f(x) dx = f(\bar x_2)(2-3) + f(\bar x_1)(1-2) + f(\bar x_0)(0-1)$$
Then clearly $\int_0^3 f(x) dx = -\int_3^0 f(x) dx$