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

[dirk_mec1]

Homework Statement

\int_0^{2018 \pi} \lvert \sin(2018x) \lvert \mbox{d}x

Homework Equations

The Attempt at a Solution

So the period is:
\frac{2 \pi}{ 2018}

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:

4 \cdot \frac{2018 \pi}{\frac{2 \pi}{ 2018} } = 2 \cdot 2018^2

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

 

[tnich]

Homework Statement

\int_0^{2018 \pi} \lvert \sin(2018x) \lvert \mbox{d}x

Homework Equations

The Attempt at a Solution

So the period is:
\frac{2 \pi}{ 2018}

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:

4 \cdot \frac{2018 \pi}{\frac{2 \pi}{ 2018} } = 2 \cdot 2018^2

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

What is $\int sin(2018x) dx$?

 

[dirk_mec1]

1/2018 * -cos(2018x) + C.

Why do you ask?

 

[willem2]

Each "hump" of the sine has an area of 2

Can you prove this?

 

[dirk_mec1]

Thanks. I understand my mistake!