Prove that $\int _0^1x^a\left(1-x\right)^bdx$ = $\int _0^1x^b\left(1-x\right)^adx$, where $a,b\in \mathbb{R}$

Question

Prove that $$\int _0^1x^a\left(1-x\right)^bdx = \int _0^1x^b\left(1-x\right)^adx$$

How can I even get started on this? I evaluate the integral with parts, but it just gets more and more tedious since I'm working with these constants here.

Answer 1

Substitute $u=(1-x)$. We then have $du=-dx$ and when $x=0$, we have $u=1$, $x=1$ gives $u=0$. Thus

$$\int_0^1x^a(1-x)^bdx=-\int_1^0(1-u)^au^bdu=\int_0^1(1-u)^au^bdu$$

No integration by parts or anything necessary, just a straight substitution.

Answer 2

Let's change variable $y=1-x$ i.e $dx=-dy$. The integral rewrites as follows

$$\int_0^1x^a(1-x)^bdx=-\int_1^0(1-y)^ay^bdy=\int_0^1(1-x)^ax^bdx$$

Answer 3

One way to answer this is to note that this is the definition of the Beta function, which can be written as $$\operatorname{B}(a+1, b+1)\\= \frac{a!b!}{(a+b+1)!}\\= \frac{b!a!}{(b+a+1)!}\\= \operatorname{B}(b+1, a+1)$$