Thread: Proof of a problem involving definate integrals?

1. Proof of a problem involving definate integrals?

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...problem.jpg

2. Originally Posted by twohaha
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?

3. Ok...

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)

4. Originally Posted by twohaha
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 $t=1-x$ then $dx=-dt$ and $\begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}$.
So $\int_0^1 {{x^a}{{\left( {1 - x} \right)}^b}dx} = \int_1^0 { - {{\left( {1 - t} \right)}^a}{t^b}dt}$

5. Originally Posted by pka
Let $t=1-x$ then $dx=-dt$ and $\begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}$.
So $\int_0^1 {{x^a}{{\left( {1 - x} \right)}^b}dx} = \int_1^0 { - {{\left( {1 - t} \right)}^a}{t^b}dt}$
What does $\begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}$.mean?

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

6. Originally Posted by twohaha
What does $\begin{array}{*{20}{c}} x&\| & 0&1 \\ \hline t&\| & 1&0 \end{array}$.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.

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