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Integral involving abs

  • Thread starter Jbreezy
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  • #1
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Homework Statement



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


Homework Equations




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
 

Answers and Replies

  • #2
33,503
5,191

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?

Homework Equations




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
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Your right its 5 sorry.
 
  • #5
33,503
5,191
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).
 
  • #6
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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?
 
  • #7
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Sure, that would work.
 
  • #8
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Thanks Mark 44.
 
  • #9
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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
 
  • #10
33,503
5,191
No. How many of those half sections are there? Hint: not 7.
 
  • #11
33,503
5,191
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
 
  • #12
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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.
 
  • #13
33,503
5,191
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.
 
  • #14
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I don't see where. I got 2sqrt2 +2sqrt2 - (0 + 1) = sqrt2 - 1
 
  • #15
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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.
 
  • #16
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No your right about the number never mind there are 8. But I can't find my mistake in my evaluation.
 
  • #17
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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.
 
  • #18
vela
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What are ##\cos \pi/4## and ##\sin \pi/4## equal to?
 
  • #19
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Both are √2/2
When you add them you get (2√2)/2 = √2
 
  • #20
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Exactly. So where did you get 2√2 that you showed in a previous post?
 
  • #21
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Oh that lol. From my tired brain. Sorry sorry. I hadn't realized it.
OK, so it is 8(sqrt2 - 1) for total area.
 
  • #22
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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.
 
  • #23
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Yeah I wanted to count that first one for some reason. I need sleep. Later dude. Thanks for the help.
 

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