If $I=\int_{-3}^{2}(|x+1|+|x+2|+|x-1|)dx$, then I equals:
Answer & explanation
Correct answer: option 1
x = -2, -1, 1 critical point
|
Case I : |
x ≤ -2 |
y = -3x - 2 |
|
Case II : |
-2 ≤ x ≤ -1 |
y = 2 - x |
|
Case III : |
-1 ≤ x ≤1 |
y = x + 4 |
|
Case IV : |
x ≥ 1 |
y = 3x + 2 |
$A_1+A_2+A_3+A_4=1(\frac{7+4}{2})+1(\frac{4+3}{2})+2(\frac{3+5}{2})+1(\frac{5+8}{2})=\frac{11+7+16+13}{2}=\frac{47}{2}$
