I am studying Apostol's Calculus Vol1. And in page 217 , I have trouble with the question 24. The problem is:
If m and n are positive integers, show that:
$\int\limits_0^1 {{x^m} \times {{(1 - x)}^n}dx} = \int\limits_0^1 {{x^n} \times {{(1 - x)}^m}dx} $

my work

I make the question into 3 different situations: m = n, m > n, m < n (the third situation is the same as the second).I cannot prove the second situation.Here is my try:
If m > n then, $\int\limits_0^1 {{x^m} \times {{(1 - x)}^n}dx} = \int\limits_0^1 {{x^n} \times {{(1 - x)}^m}dx}$ $ \Leftrightarrow $$\int\limits_0^1 {{x^n} \times {{(1 - x)}^n} \times {x^{m - n}}dx} = \int\limits_0^1 {{x^n} \times {{(1 - x)}^n} \times {{(1 - x)}^{m - n}}dx} $
And I am stuck.Any hint will be appreciate !Thanks!

  • 3
    $\begingroup$ If you substitute $t = 1-x$, what comes of that? $\endgroup$ – Daniel Fischer Aug 17 '15 at 11:34
  • $\begingroup$ Use $\displaystyle\int_a^bf(y)\ dy=\int_a^bf(a+b-y)\ dy,$ $\endgroup$ – lab bhattacharjee Aug 17 '15 at 11:55

Why not change variables: $ x^\prime = 1 - x$?

  • $\begingroup$ Yes, I got it ! Thanks! $\endgroup$ – Han Tang Aug 17 '15 at 11:59

You might be getting confused with the dummy variable $x$ in your equations. Note that if $f$ is an integrable function on $[a,b]$ then $$ \int_a^b f(t) \ dt = \int_a^b f(x) \ dx$$ This seems like a trivial point, and it is once someone has pointed it out to you, but it clears up why proving things like $$ \int_a^b f(x) \ dx = \int_{0}^{b - a} f(x + a) \ dx $$ the $x$ here is irrelevant.

Now on your question. Instead of considering both cases lets just fix $n$ and $m$. In the above example I used the transformation $u = x - a$ in order to get the integral. Can you see a similar transformation which would directly give your results? Hint: shift your $x$ a bit!


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