Let I(a,b)= int[x^a(1-x)^b]dx on bounds (0,1), where a and b are whole numbers. Note: integration by parts

a. Use substitution to show that I(a,b)=I(b,a)
b. Show that I(a,0)=I(0,a)=1/(a+1)
c. Show that I(a,b)=1/(a+1)I(a-1,b+1), for a greater than or equal to 1 and b is greater than or equal to 0.
d. Use parts b and c to calculate I(1,1) and I(3,2)

a) Let z = 1 - x.
==> x = 1 - z, dx = -dz.

So, I(a, b)
= ∫(x = 0 to 1) x^a (1 - x)^b dx
= ∫(z = 1 to 0) (1 - z)^a z^b * (-dz)
= ∫(z = 0 to 1) z^b (1 - z)^a dz
= I(b, a).
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b) I(a, 0)
= ∫(x = 0 to 1) x^a (1 - x)^0 dx
= x^(a+1)/(a+1) {for x = 0 to 1}
= 1/(a+1).

By part a, I(0, a) = I(a, 0) = 1/(a+1).
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c) Use integration by parts.
Let u = x^a, dv = (1 - x)^b dx
du = ax^(a-1) dx, dv = -(1 - x)^(b+1)/(b+1).

So, I(a, b)
= ∫(x = 0 to 1) x^a (1 - x)^b dx
= -x^a (1 - x)^(b+1) / (b+1) {for x = 0 to 1} - ∫(x = 0 to 1) [-ax^(a-1) (1 - x)^(b+1)/(b+1)] dx
= 0 + [a/(b+1)] ∫(x = 0 to 1) x^(a-1) (1 - x)^(b+1) dx
= [a/(b+1)] I(a-1, b+1).

P.S.: Did you write the formula correctly?
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d) Using parts b and c,
I(1, 1) = (1/2) * I(0, 2) = (1/2) * (1/3) = 1/6.
I(3, 2) = (3/3) * I(2, 3) = (3/3) * (2/4) * I(1, 4) = (1/2) * [(1/5) * I(0, 5)] = (1/10) * (1/6) = 1/60.

(Verified by Wolfram Alpha.)

I hope this helps!