A property of definite integrals

Question

I was studying definite integrals and there is a property which is frequently encountered: $$\int_a^b f (x) \, dx=\int_a^b f (a+b-x) \, dx$$

Well, in one of my textbooks it is written that $f (x) $ should be continuous in $(a,b) $ for this property to hold good and in another textbook no such thing is mentioned. Also, in the textbook in which the condition for $f $ to be continuous is mentioned, the fact that $f $ is continuous is not used anywhere in the proof in the book.

So what is the real thing? Does $f $ need to be continuous or not?

According to me $f $ need not be continuous because the graph of $f (x) $ and the graph of $f (a+b-x) $ are just flipped about $x=\frac {a+b}{2}$

Answer 1

Continuity is a relatively common condition needed for things to behave nicely, which is why one author included it. At least for Riemann and Lebesgue integrals, continuity is not needed for $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ to hold. My suspicion is that the property holds for most (if not all) other integrals as well.

• In the Riemann case, a the integral is typically defined in terms of tagged partitions as the maximum size of each of the partition's subintervals approaches $0$. Any such partition for the left-hand integral has a natural counterpart for the right-hand (and vice versa) by applying the $a+b-x$ transformation to the endpoints and the tag. One can use that to formally prove the two limits defining the two respective Riemann integrals are the same.
• The Lebesgue case works similarly, though instead of tagged partitions you would be working with simple functions. Mirroring the domain as you described allows the limits to be readily shown to be equal.

Answer 2

It is not necessary for $f$ to be continuous, but it is possible that an introductory textbook gave a definition of the integral that is applicable only when $f$ is continuous, in which case that may be the reason why the proposition is stated in that way.