If $I(m, n)=\int\limits_0^1 t^m(1-t)^n d t$, then the expression for $I(m, n)$ in terms of $I(m+1, n-1)$, is
Answer & explanation
Correct answer: option 2
We have,
$I(m, n)=\int\limits_0^1 t^m(1-t)^n d t$
$\Rightarrow I(m+1, n-1) =\int\limits_0^1 t^{m+1}(1-t)^{n-1} d t$
$\Rightarrow I(m+1, n-1) =\left[-\frac{t^{m+1}(1-t)^n}{n}\right]_0^1+\frac{m+1}{n} \int\limits_0^1 t^m(1-t)^n d t$
$\Rightarrow I(m+1, n-1)=\frac{m+1}{n} I(m, n)$
$\Rightarrow I(m, n)=\frac{n}{m+1} I(m+1, n-1)$