antiderivative of |sin x| on [0, 2pi] (I get 0 and 4...?)

[heartshapes]

I need to compute the antiderivative of |sin x| on [0,2pi]

I thought it was just |-cos x| evaluated from [0,2pi] so I did |-cos(2pi)|-|-cos(0)| and got 0 but when I did f[sub:3155ld3i]int[/sub:3155ld3i](|sin x|,x,0,2pi) I got 4.. help!

 

[galactus]

Re: antiderivative of |sin x|

\(\displaystyle \int_{0}^{2\pi}|sin(x)|dx=4\)

4 is correct.

 

[heartshapes]

Re: antiderivative of |sin x|

\(\displaystyle \int_{0}^{2\pi}|sin(x)|dx=4\)

4 is correct.

Thanks but what is the equation that I plug 2pi and 0 into?

Just integrate \(\displaystyle 2\int_{0}^{\pi}sin(x)dx\)

Here's the graph of |sin(x)|. See why you can do that?.

 

[heartshapes]

Re: antiderivative of |sin x|

\(\displaystyle \int_{0}^{2\pi}|sin(x)|dx=4\)

4 is correct.

Thanks but what is the equation that I plug 2pi and 0 into?

Just integrate \(\displaystyle 2\int_{0}^{\pi}sin(x)dx\)

Oh! Okay thank you. Since they are the same area value from 0 to pi and pi to 2pi I can just multiply it by 2

 

[mmm4444bot]

... when I did f[sub:sblbg0ag]int[/sub:sblbg0ag](|sin x|,x,0,2pi) I got 4 ...

Actually, your machine got 4.