How can I prove exercise 89, section 5.5 from Stewart, Calculus - Early transcendentals (7th ed)

Question

For $a$ and $b$ positive, prove that: $$\int_0^1 x^a(1-x)^b dx = \int_0^1 x^b(1-x)^a dx $$

I've tried: $$ u = 1-x, du = -dx $$

$$ x = 1 - u $$

$$ \int_1^0 (1-u)^au^b (-du) = \int_0^1 u^b (1-u)^a du $$

Now I can't think. What should I do? I have no idea.

There are multiple versions of Stewart's book (single variable/combined/early transcendentals/etc). Please learn to provide citations correctly, and provide a useful citation. — Arturo Magidin, Feb 15 '22 at 15:19

score 4 · Answer 1 · answered Feb 15 '22 at 15:09

4

You have actually finished the proof. Note that in $\int_0^1 u^b (1-u)^a du$, $u$ is just a dummy variable, so we can replace it by any other variable. Hence, $$\int_0^1 x^a(1-x)^b dx = \int_0^1 u^b (1-u)^a du = \int_0^1 x^b (1-x)^a dx.$$

answered Feb 15 '22 at 15:09

Nighty

2,152

1

Thanks! I thought I had to manipulate u and x again. Because u = 1 - x, I was replacing 1 - x. – Gabriel Bento Feb 15 '22 at 15:20

How can I prove exercise 89, section 5.5 from Stewart, Calculus - Early transcendentals (7th ed)

1 Answers1