Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

If $I=\int_{-3}^{2}(|x+1|+|x+2|+|x-1|)dx$, then I equals:

Options:

$\frac{47}{2}$

$\frac{45}{2}$

$\frac{37}{2}$

$\frac{39}{2}$

Correct Answer:

$\frac{47}{2}$

Explanation:

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

$A_1+A_2+A_3+A_4=1(\frac{7+4}{2})+1(\frac{4+3}{2})+2(\frac{3+5}{2})+1(\frac{5+8}{2})=\frac{11+7+16+13}{2}=\frac{47}{2}$