# Integral involving abs

• Jbreezy

## Homework Statement

f(x) = |sinx| and g(x)= |cosx|
I want the area of this between 0 and 2∏

## The Attempt at a Solution

So I'm just being careful. I just did the intervals and by doing g(x) - f(x) because it is the upper minus the lower. So, I got sinx + cosx between whatever interval I wanted. (∏/4), (3∏/4)...etc
But then I was like wait a minute. What is up with the abs.
So I'm just wondering about this. I feel like my integral is wrong if I don't account for it.

Thanks,
J

## Homework Statement

f(x) = |sinx| and g(x)= |cosx|
I want the area of this between 0 and 2∏
You need to be more specific. What does "this" refer to? Do you mean the area of the five regions that lie between the two curves?

## The Attempt at a Solution

So I'm just being careful. I just did the intervals and by doing g(x) - f(x) because it is the upper minus the lower. So, I got sinx + cosx between whatever interval I wanted. (∏/4), (3∏/4)...etc
But then I was like wait a minute. What is up with the abs.
So I'm just wondering about this. I feel like my integral is wrong if I don't account for it.

Thanks,
J

4 regions right? Here is the link to wolfram so you can see.
http://www.wolframalpha.com/input/?i=y+=+|cosx|+,y+=+|sinx|+
Yeah it is just the area between curves

I see five regions between 0 and 2π. The five regions are above the intervals [0, π/4], [π/4, 3π/4], [3π/4, 5π/4], [5π/4, 7π/4], and [7π/4, 2π].

You can get rid of the absolute values by using the fact that |u| = u if u ≥ 0, and |u| = -u if u < 0. You'll probably need to break up the longer subintervals above so that you can eliminate the absolute values.

For example, on [0, π], sin(x) ≥ 0, so |sin(x)| = sin(x), but on [π, 2π], sin(x) ≤ 0, so |sin(x)| = -sin(x).

Also I can just do 2 times the integral between 0 and Pi/4 then take one of the intervals in the middle and do 3 times that integral? Symmetry?

Sure, that would work.

Thanks Mark 44.

Hi again,
This is what I did. I kind of took half sections and just multiplied.

I took the integral from 0 to ∏/4 of |cosx|-|sinx|
Since both sinx and cosx are positive on this interval I got A=∫(cosx-sinx)dx [0,∏/4]
which is equal to sinx + cosx] I evaluated this at ∏/4 and 0 I got 2√2-1.
I then counted the other half sections for the area inbetween curves. I count 7 more of thease half sections between ∏/4 and 2∏. So I for total area I did 7( 2√2-1).
Yes or no?
Thanks

No. How many of those half sections are there? Hint: not 7.

Hi again,
This is what I did. I kind of took half sections and just multiplied.

I took the integral from 0 to ∏/4 of |cosx|-|sinx|
Since both sinx and cosx are positive on this interval I got A=∫(cosx-sinx)dx [0,∏/4]
which is equal to sinx + cosx] I evaluated this at ∏/4 and 0 I got 2√2-1.
Also, check your work here. It looks like you made a mistake when you evaluated sin(x) + cos(x) (which is the correct antiderivative).
I then counted the other half sections for the area inbetween curves. I count 7 more of thease half sections between ∏/4 and 2∏. So I for total area I did 7( 2√2-1).
Yes or no?
Thanks

Well I accounted for the first one. And then I count 2 between pi/4 and 3pi/4, 2 between 3pi/4 and 5pi/4, 2 between 5pi/4 and 7pi/4 and 1 for 7pi/4 and 2pi.
That is 7.

Well I accounted for the first one. And then I count 2 between pi/4 and 3pi/4, 2 between 3pi/4 and 5pi/4, 2 between 5pi/4 and 7pi/4 and 1 for 7pi/4 and 2pi.
That is 7.
No, that's eight of them altogether, not seven.

Note also that your value of 2√2 - 1 is incorrect.

I don't see where. I got 2sqrt2 +2sqrt2 - (0 + 1) = sqrt2 - 1

No, that's eight of them altogether, not seven.

Note also that your value of 2√2 - 1 is incorrect.
But I found the first area between 0 and pi/4 so I don't count that again. I have 7 more of those half sections so that is 7 times.

No your right about the number never mind there are 8. But I can't find my mistake in my evaluation.

I just did it on my calculator and with the integral option (ti 83) and I got the same thing that I got when I did it by hand.

What are ##\cos \pi/4## and ##\sin \pi/4## equal to?

Both are √2/2
When you add them you get (2√2)/2 = √2

Exactly. So where did you get 2√2 that you showed in a previous post?

Oh that lol. From my tired brain. Sorry sorry. I hadn't realized it.
OK, so it is 8(sqrt2 - 1) for total area.

That's more like it. I didn't do the integral from scratch, but your analysis on the [STRIKE]seven[/STRIKE] eight sections seems reasonable to me.

Yeah I wanted to count that first one for some reason. I need sleep. Later dude. Thanks for the help.