Question

The integral `int_0^2||x - 1| -x|dx` is equal to ______.

Options

• 1.50

• 1.60

• 1.70

• 1.80

Answer 1

The integral `int_0^2||x - 1| -x|dx` is equal to 1.50.

Explanation:

`int_0^2||x - 1| -x|dx`

= `int_0^1|-x + 1 - x|dx + int_1^2|x - 1 - x|dx`

= `int_0^1|-2x + 1|dx + int_1^2|-1|dx`

= `int_0^(1/2) + (-2x + 1)dx + int_(1/2)^1 -(-2x + 1)dx + [x]_1^2`

= `[-x^2 + x]_0^(1/2) + [x^2 - x]_(1/2)^1 + 1`

= `(-1/4 + 1/2) + 1 - 1 - 1/4 + 1/2 + 1`

= `1/4 + 1/4 + 1`

= `3/2`

= 1.50