Exercise 1542Combinatorics › Binomial coefficientsL1Added 3 Nov 2026

An integral identity

#integral#binomial#identity

Statement

Show that $\int_0^1 x^k(1-x)^{n-k}\,dx=\frac{1}{(n+1)\binom{n}{k}}$ .

Hint

Use the Beta function: $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ .

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Beta integral. $\int_0^1 x^a(1-x)^b\,dx=B(a+1,b+1)=\frac{a!\,b!}{(a+b+1)!}$ .

Apply. $a=k$ , $b=n-k$ : $\int_0^1 x^k(1-x)^{n-k}\,dx=\frac{k!(n-k)!}{(n+1)!}=\frac{1}{(n+1)\binom{n}{k}}$ .

Conclusion. The identity holds.