- #1

dirk_mec1

- 761

- 13

## Homework Statement

[tex] \int_0^{2018 \pi} \lvert \sin(2018x) \lvert \mbox{d}x [/tex]

## Homework Equations

## The Attempt at a Solution

So the period is:

[tex] \frac{2 \pi}{ 2018} [/tex]

Each "hump" of the sine has an area of 2 so if I count the number of humps I am done. In one period of an absolute sine function the area is thus 4.

So the requested area is:

[tex] 4 \cdot \frac{2018 \pi}{\frac{2 \pi}{ 2018} } = 2 \cdot 2018^2[/tex]

I am off by a factor of 2018. Where is my mistake?