Calculating Area Between Absolute Value Functions

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In summary, the problem involves finding the area between the functions f(x) = |sinx| and g(x) = |cosx| between 0 and 2π. The approach used is to subtract the upper function from the lower function and then integrate over the specified interval. The absolute values can be eliminated by breaking up the interval into smaller sections where the functions have a consistent sign. The final answer is 8(sqrt2 - 1) for the total area.
  • #1
Jbreezy
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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
 
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  • #2
Jbreezy said:

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?
Jbreezy said:

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
Your right its 5 sorry.
 
  • #5
Jbreezy said:
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
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?
 
  • #8
Thanks Mark 44.
 
  • #9
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
No. How many of those half sections are there? Hint: not 7.
 
  • #11
Jbreezy said:
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).
Jbreezy said:
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
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
Jbreezy said:
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
I don't see where. I got 2sqrt2 +2sqrt2 - (0 + 1) = sqrt2 - 1
 
  • #15
Mark44 said:
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
No your right about the number never mind there are 8. But I can't find my mistake in my evaluation.
 
  • #17
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
What are ##\cos \pi/4## and ##\sin \pi/4## equal to?
 
  • #19
Both are √2/2
When you add them you get (2√2)/2 = √2
 
  • #20
Exactly. So where did you get 2√2 that you showed in a previous post?
 
  • #21
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
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
Yeah I wanted to count that first one for some reason. I need sleep. Later dude. Thanks for the help.
 

FAQ: Calculating Area Between Absolute Value Functions

What is an integral involving abs?

An integral involving abs is a type of mathematical integral where the absolute value function (|x|) is used as part of the integrand. This means that the absolute value of the function being integrated is taken into account, resulting in a different solution than a regular integral.

Why is the absolute value function used in these integrals?

The absolute value function is used in these integrals because it allows for a more accurate representation of the area under the curve. In some cases, using the regular integral without the absolute value function may result in a negative value, which does not accurately reflect the actual area. By using the absolute value function, this issue is eliminated.

How do you solve an integral involving abs?

To solve an integral involving abs, you first need to identify the limits of integration and the integrand. Then, you can use the properties of integrals and the rules of absolute value to simplify the problem. Finally, you can use integration techniques such as substitution or integration by parts to solve the integral.

What are some real-world applications of integrals involving abs?

Integrals involving abs have many applications in physics, engineering, and economics. They are commonly used to find the displacement of an object, determine the total distance traveled, and calculate the work done by a force. In economics, they can be used to calculate consumer surplus and producer surplus.

Are there any limitations to using integrals involving abs?

One limitation of using integrals involving abs is that they may not always be applicable in certain situations. For example, if the function being integrated is discontinuous, then the solution obtained from an integral involving abs may not be accurate. Additionally, some integration techniques may not work for these types of integrals, making them more difficult to solve.

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