First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{3} | 2 - x | d x = \frac{5}{k} t h e n k =$

Moderate

A

1
B

2
C

3
D

4

Solution

$\begin{array}{l} | 2 - x | = \{\begin{cases} 2 - x, & 0 \leq x \leq 2 \\ x - 2, & 2 < x < 3 \end{cases} \\ \int_{0}^{3} | 2 - x | d x = \int_{0}^{2} (2 - x) d x + \int_{2}^{3} (x - 2) d x \end{array}$
$\begin{array}{l} = 2 x - {{\frac{x}{2}}^{2}|}_{0}^{2} + {(\frac{x^{2}}{2} - 2 x)|}_{2}^{3} \\ = 4 - 2 + \frac{9}{2} - 6 - (\frac{4}{2} - 4) \\ = \frac{5}{2} \end{array}$

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