First slide

Evaluation of definite integrals

Question

Let $f (x) = {x}$ the fractional part of $x$ then
$\int_{- 1}^{1} f (x) d x$ is equal to

Moderate

A

1
B

2
C

0
D

$1 / 2$

Solution

$f (x) = x - [x] so \int_{- 1}^{1} f (x) d x = \int_{- 1}^{0} (x + 1) d x + \int_{0}^{1} x d x$
$= 0 - (\frac{1}{2} - 1) + \frac{1}{2} = 1$

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