# How to prove that $\int_a^b f(x)dx=-\int_b^a f(x)dx$?

I came across a question on a high school level exam paper. The question reads:

Show that$$\int_a^b x^ndx=-\int_b^a x^n dx.$$

Of course, students are just expected to find the antiderivative and simply substitute the limits. However, this question drags me into deep thoughts: how do we formally define integrals where the upper limit is smaller than the lower limit?

Clearly, we should have $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$. But when I was trying to "prove" this from the definition of Riemann integrals, I find that if $$a>b$$, then dissections of the interval $$[a,b]$$ of the form $$a=x_0 is impossible. Apparently, we have not defined the integral $$\int_a^b f(x)dx$$ for $$a>b$$ in real analysis books. Lebesgue integrals don't help either, because it is integrating over sets rather than across two limits.

And here is the problem: for $$a>b$$, we have $$\int_b^a f(x)dx=\int_{[b,a]} f(x) d\mu,$$ but $$\int_a^b f(x)dx$$ is undefined.

Of course, we can say that, by definition, $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$. But somehow, I think we should not brutally define it; we should make it look more natural, i.e., to prove it under some assumptions.

• Simply, you can remove the order restriction on $(x_k)_{0\le k\le n}$. – Simply Beautiful Art Oct 5 '19 at 1:57
• Apostol uses this as a definition – Paramanand Singh Oct 5 '19 at 2:23
• The accepted answer of math.stackexchange.com/questions/261244/… has an interesting explanation. Moreover, since this is a high school exam paper, what is the relevance of Lebesgue integration? – David K Oct 5 '19 at 2:44
• see below for a short proof using FTOC – RyRy the Fly Guy Oct 5 '19 at 4:49
• if you are satisfied with your answer, then please close the post by clicking on the green check. Thanks! – RyRy the Fly Guy Oct 15 '19 at 21:33

I'd like to think that $$\int_a^b f(x)dx=\int_{\gamma} f(z)dz$$, where $$\gamma:[0,1]\to \mathbb C$$ is a curve connecting the complex numbers $$a$$ and $$b$$. This, of course, only make sense when the integral is path-independent. Let $$\delta(t)=\gamma(1-t)$$, so $$\gamma(0)=\delta(1)=a,\gamma(1)=\delta(0)=b$$. $$\int_a^b f(x)dx=\int_{\gamma} f(z)dz=\int_{[0,1]}f(\gamma(t))\dot\gamma(t) dt.$$ The last integral can be taken in the sense of either Riemann or Lebesgue. Similarly, $$\int_b^a f(x)dx=\int_{[0,1]}f(\delta(t))\dot\delta(t) dt=\int_{[0,1]}f(\gamma(1-t))(-\dot\gamma(1-t)) dt=-\int_{[0,1]}f(\gamma(t))\dot\gamma(t) dt.$$ So, $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$.

When taking a Riemann sum over an interval right to left (rather than the more common left to right), the $$\Delta x$$s are negative. This reverses the signs of all the terms of the Riemann sum (compared to what you would get for the same sum but going left to right).

We define, $$\int_a^b f = -\int_b^a f$$

There are numerous reasons why. The most basic reason is that, from the chain-rule, $$\int_a^b (f\circ g)g' = \int_{g(a)}^{g(b)} f$$ But that formula will not make sense if $$g(a) > g(b)$$, and to "correct" it, we simply define the integral by flipping the intergration.

This can be shown using the Fundamental Theorem of Calculus as follows:

Let $$F(x)$$ be a a differentiable function on $$[a,b]$$ s.t. $$F'(x)=f(x)$$. Then,

$$\int_a^b f(x)dx=F(b)-F(a)$$ and

$$\int_b^a f(x)dx=F(a)-F(b)=-[-F(a)+F(b)]=-[F(b)-F(a)]=-\int_a^b f(x)dx$$

Hence,

$$\int_b^a f(x)dx=-\int_a^b f(x)dx$$ or if we multiply both sides by $$-1,$$ $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$