Let -22 (| x|+[x x] )dx=2(3- 2)+ , where [ ] is the greatest integer function. Then ( 2 ) equals:
- 1
- 2
- 4
- 8
Answer
(2) 2
Solution
| x| and x x are even functions.
-22| x| dx=2 02 x dx=2(1- 2)
On [0, 2]: ddx(x x)= x+x x>0 (since x>-x there), so x x increases from 0 to 2 2 1.82.
Let x0 (0, 2) with x0 x0=1. Then [x x]=0 on [0, x0) and =1 on [x0, 2].
-22[x x] dx=2(2-x0)
Integral =2-2 2+4-2x0=2(3- 2)-2x0
=-2x0
2=(-2x0) (-x0)=2x0 x0=2
Answer: (2)