- #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?