If $I(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx,\ m,n0$, then $I(9,14)+I(10,13)$ is

Question

If $I(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx,\ m,n>0$, then $I(9,14)+I(10,13)$ is

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $I(m,n)$ is the Beta function and satisfies $I(m,n)=I(m+1,n)+I(m,n+1)$ (since $x^{m-1}(1-x)^{n-1}=x^{m-1}(1-x)^{n-1}[x+(1-x)]$). With $m=9,n=13$: $I(9,13)=I(10,13)+I(9,14)$. Answer (4).

Q. If $I(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx,\ m,n>0$, then $I(9,14)+I(10,13)$ is

Explanation

Reviewed by Dr. Abubacker

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