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

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

Here, I(m,n) =\(\int_0^1 t^m(1+t)n dt ,\) reduce into I(m + 1, n - 1) 
[we apply integration by parts taking (1 + t)\(^n\) as first 
and t\(^m\) as second function] 
\(\therefore \, \, \, \, \, i(m,n)=\bigg[(1+t)^n.\frac{t^{m+1}}{m+1}\bigg]_0^1 \, -\int_0^1n(1+t)^{n-1}.\frac{t^{m+1}}{m+1}dt\)
\(\, \, \, \, \, \, \, \, =\frac{2^n}{m+1}-\frac{n}{m+1} \int_0^1 (1+t)^{(n-1)},t^{m+1}dt\)
\(\therefore \, \, I(m,n) =\frac{2^n}{m+1}-\frac{n}{m+1}.I(m+1,n-1)\)