Proof of a problem involving definate integrals?

[twohaha]

How am I supposed to approach this problem?
I tried using integration by parts, but there was no way to eliminate the exponent...
Maybe I need to use the odd/even function property with integrals? I am not sure how to proceed...

 

[pka]

How am I supposed to approach this problem?
I tried using integration by parts, but there was no way to eliminate the exponent...

I doubt that any one can read what you posted.
Why not type it out?

 

[twohaha]

Ok...

If a and b are positive numbers, show that
(&int(from 0 to 1)x^a(1-x)^bdx) = (&int (from 0 to 1) x^b(1-x)^adx)

 

[pka]

If a and b are positive numbers, show that
(&int(from 0 to 1)x^a(1-x)^bdx) = (&int (from 0 to 1) x^b(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 \(\displaystyle \int_0^1 {{x^a}{{\left( {1 - x} \right)}^b}dx} = \int_1^0 { - {{\left( {1 - t} \right)}^a}{t^b}dt} \)

 

[twohaha]

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 \(\displaystyle \int_0^1 {{x^a}{{\left( {1 - x} \right)}^b}dx} = \int_1^0 { - {{\left( {1 - t} \right)}^a}{t^b}dt} \)

What does \(\displaystyle \begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}\).mean?

Is this a type of integral? My knowledge in calculus isn't very advanced...

 

[pka]

What does \(\displaystyle \begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}\).mean?

Do you understand change of variables?
As we change from x to t we have to change the limits of integration.
That is what that chart is about.