Why Riemann integral is not direction invariant

Question

Denote for $f(x)$ - some Riemann integrable function. Given a partition $\{x_0 = a, x_1, x_2, \ldots, x_n = b\}$ of $[a, b]$ we define a lower Darboux Sum $\underline{S} = \sum\limits_{i}\min\limits_{x \in [x_{i-1}, x_i]}f(x)(x_i - x_{i-1})$ and respectively an upper Darboux Sum $\bar{S}$.

I see this as a sum of rectangle areas, where sign is only inherited from the value of $f(x)$, while $(x_i-x_{i-1})$ is just the length of the rectangle basis.

What bothers me, is that

the length of the rectangle basis is independent of the direction I'm looking on it (from right or from left), then why does the sign of the integral is not independent of the direction?

i.e., why $$\int\limits_a^b f(x) \mathrm{d}x = -\int\limits_b^a f(x) \mathrm{d}x$$

Answer 1

For $a<b$, we define $\int_a^b f(x)\textrm{d}x$ as the supremum of the lower Darboux sums. On the other hand, we define $\int_b^a f(x)\textrm{d}x=-\int_a^b f(x)\textrm{d}x$ in order to make the identity

$$ \int_a^b f(x)\textrm{d}x+\int_b^c f(x)\textrm{d}x=\int_a^c f(x)\textrm{d}x $$ hold no matter the relation of $a,b$ and $c$.

Alternatively, when integrating from $a$ to $b,$ you observe a partition $a=x_0<x_1<...x_n<b$ and observe the Riemann sums $\sum_{j=1}^n f(\xi) (x_j-x_{j-1})$. The analogous construction when integrating from $b$ to $a$ would be to observe a partion $b=x_0>x_1>...>x_n=a$ and the corresponding Riemann sums $\sum_{j=1}^n f(\xi) (x_j-x_{j-1})$, and here, the $x_j-x_{j-1}$ is a negative increment. This is another way to intuit that this convention might be useful.