Question

interesting integral:

Answer 1

part 1

let\[I(a,b) = \int\limits^1_0x^a(1-x)^bdx\]

show that
\[I(a,b) = I(b,a)\]

Answer 2

That reminds me of an exercise in Stewart's.

Answer 3

part 2

show that
\[I(a,b) = I(a+1,b) + I(a,b+1)\]

Answer 4

part 3

show that

\[(a+1)I(a, b) = bI(a+1, b−1)\]

when a and b are positive.

hence evaluate I(a,b) when a and b are positive integers

Answer 5

For part I, let u=1-x:
\[\int_0^1x^a(1-x)^b\ dx=-\int_1^0(1-u)^au^b\ du=\int_0^1 (1-x)^ax^b\ dx\]

Answer 6

yeeeah

Answer 7

I'm a little stuck on part 2. Does it require integration by parts?

Answer 8

My bad, it doesn't require integration by parts.
\[\int_0^1 x^{a+1}(1-x)^b\ dx+\int_0^1 x^a(1-x)^{b+1}\ dx\\
=\int_0^1 x^{a+1}(1-x)^b+x^a(1-x)^{b+1}\ dx\\
=\int_0^1 x^a(1-x)^b(x+1-x)\ dx\\
=\int_0^1 x^a(1-x)^b\ dx\]

Answer 9

Part 3 is straightforward integration by parts:
\[\int_0^1 x^{a+1} b(1-x)^{b-1}\ dx=\left .-x^{a+1}(1-x)^b\right |_0^1+\int_0^1 (a+1)x^a(1-x)^b\ dx\\
b\int_0^1 x^{a+1} (1-x)^{b-1}\ dx=(a+1)\int_0^1 x^a(1-x)^b\ dx\]

Answer 10

correct!

Answer 11

Reminds me of the Beta function

Answer 12

i hadn't come across the beta function before, very interesting! thanks

Answer 13

reading this now http://en.wikipedia.org/wiki/Beta_function

Answer 14

I can't decrease a and b at the same time. :(

Answer 15

do you need to? if we can decrease one to zero or one, the integral is easy