Definite Integration Question 83
Question: $ \int _{-2}^{2}{| [x] |dx=} $
[EAMCET 2003]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int _{-2}^{2}{|[x]|}dx=\int _{-2}^{-1}{|[x]|dx+\int _{-1}^{0}{|[x]|dx+\int_0^{1}{|[x]|dx|+\int_1^{2}{|[x]|dx}}}} $
$ =\int _{-2}^{-1}{2dx}+\int _{-1}^{0}{1dx+\int_0^{1}{0dx+}}\int_1^{2}{1dx} $
$ =2[x] _{-2}^{-1}+[x] _{-1}^{0}+0+[x]_1^{2} $
$ =2(-1+2)+(0+1)+(2-1)=2+1+1=4. $
BETA
