Very well done, good tutorial. For completeness, one should include calculations for the special cases of n=-1 and n=-2 since the initial question did not exclude these values which make the general answer undefined.
Another interesting problem is the generalisation : x^p(1-x)^q for natural p and q. Edit : Spoilers - (You can solve this entire class of integrals for a fixed p+q)
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@@Gala_XY it reminds me of a warm summer
1. Solution
Integration by parts with
u=x , dv = (1-x)^{n}dx
2. Solution
Substitution
u = 1-x
Very well done, good tutorial.
For completeness, one should include calculations for the special cases of n=-1 and n=-2 since the initial question did not exclude these values which make the general answer undefined.
Another interesting problem is the generalisation : x^p(1-x)^q for natural p and q.
Edit : Spoilers -
(You can solve this entire class of integrals for a fixed p+q)
Isn't it the beta function?
Beta function
Instead of suppose, you could have directly used the property f(x) = f(a-x)
Fairly easy integral
Solution:
1
∫x*(1-x)^n*dx =
0
---------------------
Substitution:
u = 1-x x = 1-u du = -dx dx = -du
upper limit = 1-1 = 0 lower limit = 1-0 = 1
---------------------
0 1 1
= -∫(1-u)*u^n*du = ∫[u^n-u^(n+1)]*du = [u^(n+1)/(n+1)-u^(n+2)/(n+2)]
1 0 0
= [1^(n+1)/(n+1)-1^(n+2)/(n+2) = 1/(n+1)-1/(n+2)
= [1*(n+2)-1*(n+1)]/[(n+1)*(n+2)] = [n+2-n-1]/[(n+1)*(n+2)]
= 1/[(n+1)*(n+2)]
Fairly easy integral