I doubt that any one can read what you posted.How am I supposed to approach this problem?
I tried using integration by parts, but there was no way to eliminate the exponent...
Let \(\displaystyle t=1-x\) then \(\displaystyle dx=-dt\) and \(\displaystyle \begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}\).If a and b are positive numbers, show that
(&int(from 0 to 1)xa(1-x)bdx) = (&int (from 0 to 1) xb(1-x)adx)
Let \(\displaystyle t=1-x\) then \(\displaystyle dx=-dt\) and \(\displaystyle \begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}\).
So \(\displaystyle \int_0^1 {{x^a}{{\left( {1 - x} \right)}^b}dx} = \int_1^0 { - {{\left( {1 - t} \right)}^a}{t^b}dt} \)
Do you understand change of variables?What does \(\displaystyle \begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}\).mean?