Why does an integral change signs when flipping the boundaries? Let us define a very simple integral:


$f(x) = \int_{a}^{b}{x}$ 


for $a,b\ge 0$.

Why do we have the identity $\int_{a}^{b}{x} = -\int_{b}^{a}{x}$?

I drew the graphs and thought about it but to me integration, at least in two-dimensions, is just taking the area underneath a curve so why does it matter which direction you take the sum?
Here's another intuitive justification. The obvious graphical intuition says that when $a \leq b \leq c$, then $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$. If we want this formula to hold for arbitrary $a,b,c$, then we should be able to take $a=c$, so that $\int_a^b f(x) dx + \int_b^a f(x) dx = \int_a^a f(x) dx$. But $\int_a^a f(x) dx = 0$, so if we want this formula to hold, we need $\int_a^b f(x) dx = -\int_b^a f(x) dx $.
The change in $x$, $dx$ becomes negative. So the integral sums over negative quantities.

This is useful for when both coordinates of a curve are functions, and you find the area under the curve by $\int_{t_1}^{t_2}y(t) \, dx(t)$. When the $x(t)$ path travels in the negative direction, $dx(t)$ is negative and so the integral decreases.