Question

If $I(m,n)={\int }_0^1{t}^m{(1+t)}^{n}dt$ then $I(m,n)$ is equal to

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

Answer 1

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

$I(m,n)={\int }_0^1{t}^m{(1+t)}^{n}dt$
$={[{(1+t)}^n.\frac{t^{m+1}}{m+1}]}_0^1-{\int }_0^{1}n{(1+t)}^{n-1}.\frac{t^{m+1}}{m+1}dt$
$=\frac{2n}{m+1}-\frac{n}{m+1}{\int }_0^1{(1+t)}^{n-1}{t}^{m+1}dt$
$\therefore I(m,n)=\frac{2^n}{m+1}-\frac{n}{m+1}.I(m+1,n-1)$