∫−22 |[x]|dx is equal to
1
2
3
4
∫−22 |[x]|dx=∫−2−1 |[x]|dx+∫−10 |[x]|dx+∫01 |[x]|dx+∫12 |[x]|dx=∫−2−1 2dx+∫−10 1dx+∫01 0dx+∫12 1dx=2[x]−2−1+[x]−10+0+[x]12=2(−1+2)+(0+1)+(2−1)=2+1+1=4