Definite Integration Question 474
Question: The value of the integral $ I=\int _{0}^{1}{x{{(1-x)}^{n}}dx} $ is
[AIEEE 2003]
Options:
A) $ \frac{1}{n+1} $
B) $ \frac{1}{n+2} $
C) $ \frac{1}{n+1}-\frac{1}{n+2} $
D) $ \frac{1}{n+1}+\frac{1}{n+2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_0^{1}{x{{(1-x)}^{n}}dx} $
$ -I=\int_0^{1}{-x{{(1-x)}^{n}}dx=\int_0^{1}{(1-x-1){{(1-x)}^{n}}dx}} $
$ =\int_0^{1}{{{(1-x)}^{n+1}}dx-\int_0^{1}{{{(1-x)}^{n}}dx}} $
$ =[ \frac{{{(1-x)}^{n+2}}}{-(n+2)} ]_0^{1}-[ \frac{{{(1-x)}^{n+1}}}{-(n+1)} ]_0^{1}=\frac{1}{n+2}-\frac{1}{n+1} $
$ \Rightarrow I=\frac{1}{n+1}-\frac{1}{n+2}. $
BETA
