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I confused for calculating $$\int_{0}^{1}x^k(1-x)^{n-k}dx$$
one solution that I guess is: $$1^{n}=(x+1-x)^{n}=\binom{n}{k}x^{k}(1-x)^{n-k}$$
so $$x^{k}(1-x)^{n-k}=\frac{1}{\binom{n}{k}}$$ finally $$\int_{0}^{1}x^k(1-x)^{n-k}dx=\int_{0}^{1}\frac{1}{\binom{n}{k}}dx=\frac{(n-k)!k!}{n!}$$

but I saw elsewhere that the answer is $$\frac{(n-k)!k!}{(n+1)!}$$
I confused which one is correct??!!

AMIR 1996
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    You forget a sum: $$1^n=(x+1-x)^n=\sum_{k=0}^n{n\choose k} x^k(1-x)^{n-k}.$$ – mickep Jan 02 '16 at 09:49
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    To pursue on your idea, introduce some parameter $t>1$ and consider $$\int_0^1(tx+1-x)^ndx=\int_0^1\sum_{k=0}^n{n\choose k}t^kx^k(1-x)^{n-k}dx=\sum_{k=0}^n{n\choose k}I_{k,n}t^k,$$ where $I_{k,n}$ denotes the integral you are interested in. Then the change of variable $u=tx+1-x$ shows the LHS is $$\int_1^tu^n\frac{du}{t-1}=\frac1{n+1}\frac{t^{n+1}-1}{t-1}=\frac1{n+1}\sum_{k=0}^nt^k,$$ hence, by identification, $${n\choose k}I_{k,n}=\frac1{n+1},$$ end of the proof. – Did Jan 03 '16 at 08:42

2 Answers2

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For each $k$, define $I_k$ as $I_k=\int_0^1x^k(1-x)^{n-k}dx$.

Using integration by parts, you'll get $I_{k+1}=\frac{k}{n-k}I_k$. Using the initial condition $I_0=\int_0^1(1-x)^ndx$ conclude.

sinbadh
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I'll be using beta function to sove the problem.

The integral:$$\large \int_{0}^{1}x^k(1-x)^{n-k}dx$$

$$\large =B(k+1,n-k+1)$$

where, B(x,y) is beta function.

$$\large =\frac { \Gamma \left( k+1 \right) \Gamma \left( n-k+1 \right) }{ \Gamma \left( n+2 \right) } $$

$$\large =\frac { k!\left( n-k \right) ! }{ \left( n+1 \right) ! } $$

Aditya Kumar
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  • Why do you delete your answer, and then add one that is the same? – mickep Jan 03 '16 at 13:44
  • This is more or less equivalent to saying that one knows the answer hence one knows the answer. – Did Jan 04 '16 at 12:04
  • Certainly not. One can definitely answer this question in a way which is not equivalent to "one knows the answer hence one knows the answer" (see my comment on main for an option). – Did Jan 04 '16 at 12:15
  • Huh? No idea how your last comment applies to the question, to your answer or to the point I made. Care to explain? – Did Jan 04 '16 at 12:19
  • Not in general, of course, this depends on the cases: in the present case, yes this is what you did (as you admitted). Sorry but why are we even discussing this? – Did Jan 04 '16 at 12:25