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

Question

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.

Answer 1

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.

Answer 2

You can (often) think of that area for $a < b$ and positive $f$ as what's on your left as you move from $a$ to $b$, up to $f$, back along the graph then down to $a$.

In that model areas on your right are negative. That gives the correct answer when $f$ takes on negative values, and it reverses sign when you reverse direction by swapping $a$ and $b$.

Answer 3

This is because the integral must satisfy additivity w.r.t. integrals, which implies $$\int_a^b f(x)\,\mathrm d x+\int_b^a f(x)\,\mathrm d x=\int_a^a f(x)\,\mathrm d x=0.$$

Answer 4

If $f(t)$ is your velocity at time $t,$ then $\int_{t_0}^{t} f(t)\,dt$ is the amount of distance you must add to your position at time $t_0$ to get your position at time $t$. This wouldn't be true for $t<t_0$ if we didn't define it this way. In particular, we'd want $\int_{t_0}^{t_1} f(t)\,dt=-\int_{t_1}^{t_0} f(t)\,dt$ if we wanted this to work property.

Answer 5

We can state that if a < b, then the length of the interval [a, b], $L = b - a$ and what the define integral make is suming all $\delta x$ from a to b, of course $L=\int_{a}^{b}dx=b-a$, the integral is the the interval length, if you compute it backwards $-L=-\int_{b}^{a}dx=a-b=-(b-a)$. Moreover in the Riemann sum partition's definition you have a < b and $P=\{a=x_0<x_1<...<x_n=b\}$ to define $\int_{a}^{b}dx$ so if you take the integral from b to a, what your are doing is suming up to the negative side since you start from b but you sum negative infinitesimals with respect to b.

Answer 6

$\int_a^bf(x)dx=-\int_b^af(x)dx$ in general... just look at the definition of the integral as a limit of Riemann sums (we get the exact same sums except moving in the reverse direction from $b$ to $a$)

Consider, for instance, an increasing $f$ and a partition of $[a,b]$ into one interval... You get a lower sum $f(a)(b-a)$ which corresponds to a lower sum $f(a)(a-b)$ for the second integral. In this way you can pair off the terms of the approximating sums...

As we are dealing for the most part with "nice" functions, i.e. integrable, we can break the function into increasing and decreasing parts, and then add up and take limits...

Alternatively, and this is sort of "hand waving", we can replace $dx$ by $-dx$ in moving from $b$ to $a$...