Q:

Why is $\int_a^b f(x)dx$ defined as $-\int_b^a f(x)dx$ for $b<a$?

I know with this definition "mathematics doesn't break" (sorry I don't know how to express it better, I mean that calculations and proofs work out), but is there a way you can interprete it? I figured that visually it would mean that $\int_a^b f(x)dx$ is equal to the area above the graph of f mirrored about the x-axis in the intervall $(b,a)$, which is not very insightful to me. How can you make sense of it? Thanks in advance. 

A:

You're right, the “area metaphor” for the integral doesn't justify this convention that well.  Negative area is kind of dubious.
You might try instead the “displacement metaphor” for the integral.  If $v(t)$ is the velocity of a particle at time $t$, then $\int_{a}^b v(t)\,dt$ is the displacement between the particle at time $a$ and time $b$.  I say displacement instead of distance to allow for direction.  If positive displacement were measured to the right, then negative displacement would go to the left.
What would $\int_b^a v(t)\,dt$ be?  One way to think about it is the same displacement result, but instead running time backwards from $b$ to $a$.  And the displacement from $t=b$ to $t=a$ (backwards) should be the opposite of the displacement from $t=a$ to $t=b$.  Meaning, same magnitude but opposite sign.

