Definite Integration Question 77
Question: The value of $ I=\int _{0}^{1}{x| x-\frac{1}{2} |dx} $ is
[UPSEAT 2003]
Options:
A) 1/3
B) 1/4
C) 1/8
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_0^{1}{x| x-\frac{1}{2} |dx} $
$ =-\int_0^{1/2}{x( x-\frac{1}{2} )+\int _{1/2}^{1}{x( x-\frac{1}{2} )}dx} $
$ =\int_0^{1/2}{( \frac{1}{2}x-x^{2} )dx+\int _{1/2}^{1}{( x^{2}-\frac{1}{2}x )dx}} $
$ =( \frac{x^{2}}{4}-\frac{x^{3}}{3} )_0^{1/2}+( \frac{x^{3}}{3}-\frac{x^{2}}{4} ) _{1/2}^{1} $
$ =( \frac{1}{16}-\frac{1}{24} )+( \frac{1}{3}-\frac{1}{4}+\frac{1}{16}-\frac{1}{24} )=\frac{1}{8}. $
BETA
