# integral abs sinx [0,3pi/2]

• Jul 28th 2010, 07:08 PM
kensington
integral abs sinx [0,3pi/2]
integral abs sinx [0,3pi/2]

would this be -cosx(abs sinx/sinx)?
• Jul 28th 2010, 07:15 PM
mr fantastic
Quote:

Originally Posted by kensington
integral abs sinx [0,3pi/2]

would this be -cosx(abs sinx/sinx)?

Draw the graph, then set up and evaluate the required integrals.
• Jul 28th 2010, 10:02 PM
autumn
You cannot ignore the absolute value.
Break up the integral into the parts where sin x is positive and negative and flip the negative part over.
Then you can integrate.
• Jul 29th 2010, 05:05 AM
kensington
what do you mean flip the negative part over?
• Jul 29th 2010, 05:16 AM
Vlasev
Quote:

Originally Posted by kensington
what do you mean flip the negative part over?

This means that $\displaystyle |sin(x)|= - sin(x)$
• Jul 29th 2010, 06:25 AM
kensington
Quote:

Originally Posted by Vlasev
This means that $\displaystyle |sin(x)|= - sin(x)$

soo then is it

abs sinx= sinx if x=> 3pi/2
=-sinx if x< 3pi/2

so then integral from 0 to pi sinx+ integral from pi to 3pi/2 -sinx
which is then [-cosx]+ [cosx]
[(-cos0)-(-cospi)]+ [(cospi)-(cos3pi/2)]
[-1-(1)] + [-1-(-1/2)]
-2-1/2
=-2.5??

Would that be correct?
• Jul 29th 2010, 06:58 AM
Plato
This may surprise you but: $\displaystyle \int_0^{\frac{{3\pi }} {2}} {\left| {\sin (x)} \right|dx} = 3\int_0^{\frac{\pi } {2}} {\sin (x)dx}$.

Think about it after looking at the graph.
• Jul 29th 2010, 07:06 AM
kensington
Quote:

Originally Posted by Plato
This may surprise you but: $\displaystyle \int_0^{\frac{{3\pi }} {2}} {\left| {\sin (x)} \right|dx} = 3\int_0^{\frac{\pi } {2}} {\sin (x)dx}$.

Think about it after looking at the graph.

so then would it be
3 [[(-cos0)-(-cospi/2)]+ [(coso)-(cospi/2)]]?
• Jul 29th 2010, 07:08 AM
Plato