Question

If $I(m,n)={\int }_0^1{t}^m{(1-t)}^{n}dt$, then the expression for $I(m,n)$ in terms of $I(m+1,n-1)$ is

• A. $\frac{2^n}{m+1}+\frac{n}{m+1}$ $I(m+1,n-1)$
• B. $\frac{n}{m+1}$ $I(m+1,n-1)$
• C. $\frac{2^n}{m+1}-\frac{n}{m+1}$ $I(m+1,n-1)$
• D. $\frac{m}{m+1}$ $I(m+1,n-1)$

Answer 1

The correct option is A $\frac{2^n}{m+1}+\frac{n}{m+1}$ $I(m+1,n-1)$

Here, $I(m,n)={\int }_0^2{t}^m(1-t{)}^{n}dt,$ reduce into $I(m+1,n-1)$

We apply integgration by parts taking $(1+t{)}^n$ as first and $t^m$ as second function.

$\Rightarrow$ $I(m,n)={\left[\right(1-t{)}^n\times \frac{t^{m+1}}{m+1}]}_0^1-{\int }_0^{1}n(1-t{)}^{n-1}\times \frac{t^{m+1}}{m+1}dt$

$\Rightarrow$ $\frac{2^n}{m+1}+\frac{n}{m+1}{\int }_0^1(1-t{)}^{(n-1)}{t}^{m+1}dt$

$\therefore$ $I(m,n)=\frac{2^n}{m+1}+\frac{n}{m+1}.I(m+1,n-1)$