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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 $a=x_0<x_1<...<x_n=b$ 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.

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  • $\begingroup$ Simply, you can remove the order restriction on $(x_k)_{0\le k\le n}$. $\endgroup$ – Simply Beautiful Art Oct 5 '19 at 1:57
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    $\begingroup$ Apostol uses this as a definition $\endgroup$ – Paramanand Singh Oct 5 '19 at 2:23
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    $\begingroup$ 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? $\endgroup$ – David K Oct 5 '19 at 2:44
  • $\begingroup$ see below for a short proof using FTOC $\endgroup$ – RyRy the Fly Guy Oct 5 '19 at 4:49
  • $\begingroup$ if you are satisfied with your answer, then please close the post by clicking on the green check. Thanks! $\endgroup$ – RyRy the Fly Guy Oct 15 '19 at 21:33
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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, $\int_a^b f(x)dx=-\int_b^a f(x)dx$.

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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).

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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.

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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 $-1,$ $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$

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