Question

If $I(m,n)={\int }_{-0}^1{t}^m(1+t{)}^{n}dt$, then 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

Answer: (A)

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

Here, $I(m,n)={\int }_0^1{t}^m(1+t{)}^{n}dt$
[We apply integration by parts, taking $(1+n)$ as first $t^m$ as second function]
$I(m,n)={\left[\right(1+t{)}^n\cdot \frac{t^{m+1}}{m+1}]}_0^1-{\int }_0^{1}n(1+t{)}^{n-1}\cdot \frac{t^{m+1}}{m+1}dt$
$=\frac{2^n}{m+1}-\frac{n}{m+1}{\int }_0^1(1+t{)}^{n-1}\cdot t^{m+1}dt$
Therefore, $I(m,n)=\frac{2^n}{m+1}-\frac{n}{m+1}\cdot I(m+1,n-1)$.