What is the mistake in calculating the integral of the absolute sine function?

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dirk_mec1
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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?
 
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dirk_mec1 said:

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?
What is ##\int sin(2018x) dx##?
 
dirk_mec1 said:
Each "hump" of the sine has an area of 2
Can you prove this?