First slide

Evaluation of definite integrals

Question

The value of $\int_{- 2}^{2} | 1 - x | d x$ is

Moderate

A

2
B

0
C

4
D

5

Solution

$\begin{array}{l} | 1 - x | = \{\begin{cases} 1 - x & , x \leq 1 \\ x - 1 & , x > 1 \end{cases} so \\ \begin{aligned} \int_{- 2}^{2} | 1 - x | d x & = \int_{- 2}^{1} | 1 - x | d x + \int_{1}^{2} | 1 - x | d x \\ = \int_{- 2}^{1} (1 - x) d x + \int_{1}^{2} (x - 1) d x \\ = 5 . \end{aligned} \end{array}$

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