Definite Integration Question 359
Question: If $ l(m,n)=\int_0^{1}{t^{m}{{(1+t)}^{n}}dt,} $ then the expression for $ l(m,n) $ in terms of $ l(m+1,n-1) $ is
[IIT Screening 2003]
Options:
A) $ \frac{2^{n}}{m+1}-\frac{n}{m+1}l(m+1,n-1) $
B) $ \frac{n}{m+1}l(m+1,n-1) $
C) $ \frac{2^{n}}{m+1}+\frac{n}{m+1}l(m+1,n-1) $
D) $ \frac{m}{n+1}l(m+1,n-1) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ l(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{2^{n}}{m+1}-\frac{n}{m+1}l(m+1,n-1) $ .
BETA
