First slide

Evaluation of definite integrals

Question

$\int_{- 1}^{1} | 2 x - 1 | d x$ is equal to

Moderate

A

$- 2$
B

$- 1 / 2$
C

$1 / 2$
D

$5 / 2$

Solution

$\int_{- 1}^{1} | 2 x - 1 | d x = \int_{- 1}^{1 / 2} | 2 x - 1 | d x + \int_{1 / 2}^{1} | 2 x - 1 | d x$
$= \int_{- 1}^{1 / 2} (- (2 x - 1)) d x + \int_{1 / 2}^{1} (2 x - 1) d x$
$= {[x - x^{2}]}_{1}^{1 / 2} + {[x^{2} - x]}_{1 / 2}^{1}$
$= (\frac{1}{2} - \frac{1}{4}) - (- 1 - 1) + (1 - 1) - (\frac{1}{4} - \frac{1}{2}) = \frac{5}{2}$

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