Hi, can someone help me with this:

If m and n are positive integers, show that

int_{0}^{1} (x^m)(1-x)^n dx (Thats the integral from 0 to 1)

is equal to

int_{0}^{1} (x^n)(1-x)^m dx.

Thanks a bunch.

Printable View

- May 2nd 2007, 09:51 PMRecklessidproof of integral
Hi, can someone help me with this:

If m and n are positive integers, show that

int_{0}^{1} (x^m)(1-x)^n dx (Thats the integral from 0 to 1)

is equal to

int_{0}^{1} (x^n)(1-x)^m dx.

Thanks a bunch. - May 2nd 2007, 10:16 PMecMathGeek
INT{0,1} x^m*(1 - x)^n dx = INT{0,1} x^n*(1 - x)^m dx

LHS: INT{0,1} x^m*(1 - x)^n dx

Let u = 1 - x <--> du = -1 dx --> dx = -1 du

The limits of integration become:

x = 0 --> u = 1 - 0 = 1

x = 1 --> u = 1 - 1 = 0

- INT{1,0} (1 - u)^m*u^n du

Switching the limits of integration back to {0,1} reverses the sign of the integration:

INT{0,1} u^n*(1 - u)^m du

Setting the LHS and RHS equal, we see that:

INT{0,1} u^n*(1 - u)^m du = INT{0,1} x^n*(1 - x)^m dx

Since both sides are the integration of the same function (though in terms of seperate variables) with the same limits of integration, the integrations are the same. The variable used in integration does not matter so long as the function and the limits are the same. - May 3rd 2007, 05:23 AMThePerfectHacker
They are the same because they are equal to Beta function.

- May 3rd 2007, 06:33 AMecMathGeek
- May 3rd 2007, 08:59 AMThePerfectHacker
- May 3rd 2007, 09:23 AMecMathGeek