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Area under a graph

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Homework Statement



Please refer to the image attached. Part (ii)

Homework Equations


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The Attempt at a Solution



My workings are in the second image. I don't understand why is it wrong.

Since the graph is well below the x-axis, if I integrate it do I get infinity?

(The x-coordinates of P and Q are -2 and 4)
 

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  • #2
Simon Bridge
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Since the graph is well below the x-axis, if I integrate it do I get infinity?
Not if you set it up right.
Areas below the x axis normally come out negative. You need to decide how to divide the region up so you can integrate in several stages.
 
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  • #3
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Alternatively, think of a way you could translate each of the graphs such that the bounded area is then totally above the x-axis, thus eliminating the confusion of having to deal with "negative areas".
 
  • #4
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Homework Statement



Please refer to the image attached. Part (ii)

Homework Equations


None


The Attempt at a Solution



My workings are in the second image. I don't understand why is it wrong.

Since the graph is well below the x-axis, if I integrate it do I get infinity?

(The x-coordinates of P and Q are -2 and 4)
If you just look at your picture you can see the area involved is not infinite. To get an integral to be ##\infty## you need a function which becomes either very large or very small (negative numbers). That isn't happening in the area of your interest.
 
  • #5
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coconut62,
Your integral is not set up correctly. In the first line of your work you have three integrals, and one of them is incorrect. Using vertical strips, which you did, all you need is one integral. Throughout the interval [-2, 4], the line is always above the parabola, and the height of a typical area element is (2x + 4) - (x2 - 4). Just integrate this expression and you should get the right answer.
 
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  • #6
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Not if you set it up right.
Areas below the x axis normally come out negative.
An area, wherever it is, should not come out negative. If it does, you haven't set up the integral correctly. If you are integrating with respect to x (using vertical strips), the height of each strip should be <yupper - ylower>. If you are integrating with respect to y (using horizontal strips), the width of each strip should be <xright - xleft>.
You need to decide how to divide the region up so you can integrate in several stages.
 
  • #7
Simon Bridge
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An area, wherever it is, should not come out negative. If it does, you haven't set up the integral correctly.
I think we are using different ideas about what an area is: the area under a velocity-time graph is the displacement and displacements can be negative.

Reversing the order of the integral limits just to make that instance positive will make other instances negative - or introduce inconsistencies into the model.
 
  • #8
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When you are looking for an area, as opposed to an integral, you have to do your computation in at least two steps. The first is to compute the integral for those values of f that are positive. Then you compute separately an integral for those values of f that are negative. Each of these computations could involve splitting up the integral still further.

Once you have the positive and negative values, say A and B respectively, your area is A - B.
 
  • #9
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I think we are using different ideas about what an area is: the area under a velocity-time graph is the displacement and displacements can be negative.
In that case you're not really talking about "area" as the dimensions would be distance/time * time, and distance can be signed. If we're talking about area in the usual geometric sense, area can't be negative.

Also, "under a graph" doesn't make sense for the intervals where the graph goes negative. Here the region under the graph would be infinite in size. A more precise description would describe the region as between the graph and the horizontal axis.

Here's an example: f(x) = sin(x) on [0, ##2\pi##].
If we want the area of the region bounded by f and the x axis, the integral would be
$$\int_0^{\pi} [sin(x) - 0]dx + \int_{\pi}^{2\pi} [0 - sin(x)]dx = 2 + 2 = 4$$

OTOH, if f represents the velocity, then integrating the velocity gives us the displacement.
$$\int_0^{2\pi} sin(x)dx = 0$$
Reversing the order of the integral limits just to make that instance positive will make other instances negative - or introduce inconsistencies into the model.
Getting back to the context of this thread (area), I don't at all suggest that we change from the usual order of integration (i.e., left to right or from bottom to top).
 
  • #10
Simon Bridge
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Yes I figured we were not talking about the same concept of "area".
You are restricting the term to refer to the amount of a geometric surface.
You'd probably say that "yeah, of course, that's what area means."

See also, discussions:
https://www.physicsforums.com/showthread.php?t=370338
https://www.physicsforums.com/showthread.php?t=292737

And this is fairly representative of calc classes:
http://tutorial.math.lamar.edu/Classes/CalcI/AreaProblem.aspx

For the area between two curves where f(x)>g(x) in the regeon a<x<b we have:

$$A=\int_a^bf(x)dx-\int_a^bg(x)dx = A_f-A_g$$ ... notice that this returns a positive area even if both functions are completely negative and their integrals (areas) are negative.

Technically all integration is finding the area between two curves - it's just that we teach g(x)=0 without stating that when you are just starting out. To make this work, some parts of the designated area must be given a negative sign.
 
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  • #11
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For the area between two curves where f(x)>g(x) in the regeon a<x<b we have:

$$A=\int_a^bf(x)dx-\int_a^bg(x)dx = A_f-A_g$$ ... notice that this returns a positive area even if both functions are completely negative and their integrals (areas) are negative.
Since f(x) > g(x) on the interval, then f(x) - g(x) > 0. So you can write the integral that represents the area between the two curves as
$$A=\int_a^b [f(x) - g(x)]dx$$
Then you don't have to resort to talking about negative areas.




Technically all integration is finding the area between two curves - it's just that we teach g(x)=0 without stating that when you are just starting out. To make this work, some parts of the designated area must be given a negative sign.
Not if you pay attention to which bounding function is larger on a given interval.
 
  • #12
Simon Bridge
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I didn't expect to convince you. (Did you also check the links?)

But now, at least, I know you are aware of other ways of looking at the concept of area than what you have expounded.
You may disagree with them - that does not make them incorrect.

The area under the x-axis being negative is the standard approach taken in text books. It doesn't stop there of course - any simple definitive statement would be misleading.
Maybe you'd say the standard approach is incorrect in some way, you can take it up with the authors, but the student still has to deal with it.
 
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  • #13
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I looked at a couple of the links in post #10. I agree with HallsOfIvy, that "area" is just one application of the definite integral, and that "area" is nonnegative. It's been a while since I taught calculus (about 15 years), but I don't remember the text we used (Thomas-Finney) or others that talked about area being negative or describing regions below the x-axis as having "negative area." Can you cite any textbooks that use what you describe (negative area) as their standard approach?

As I said before, the whole thing becomes moot if you just keep track of which function is larger on any given interval.
 
  • #14
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We're off-topic here and haven't heard from the OP since post #1. Let's agree to disagree on this negative area thing.
 
  • #15
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We're off-topic here and haven't heard from the OP since post #1. Let's agree to disagree on this negative area thing.
It's all terminology anyway. If all of us (possibly excepting the OP) worked this problem, I'm sure we would all get the same answer.
 
  • #16
Simon Bridge
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Mark44 said:
Can you cite any textbooks that use what you describe (negative area) as their standard approach?
Sure: Calculus by Gilbert Strang (2005) uses the concept repeatedly. Noteably in ch7. (MIT have an online study guide and an instructors manual for it.)

Edwards and Penny (1986), Calculus and Analytic Geometry 2nd Ed - takes trouble to distinguish between "negative area" and "the negative of the area" for different contexts (i.e. p279)...

Both those are college level - HS texts are more likely to get sloppy with terminology.

I know where you are coming from - the geometric area has dimensions of L^2 and can only be negative in the sense of removing part of the surface. (i.e. "negative space" in art) A negative integral for geometric area is an artifact of the coordinate system.

As I said before, the whole thing becomes moot if you just keep track of which function is larger on any given interval.
I believe we agree there:)

When doing integrations, you have to bear in mind what it is that you are trying to find.
If you do an integration and get a negative value out the end - it is not necessarily invalid. A negative outcome may make sense for the problem. I don't think you disagree about this.

brmath said:
It's all terminology anyway.
Yeah - the disagreement is over nomenclature.

In this case, the final area should be positive ... as in, how much carpet would cover a floor with that shape.

We're off-topic here and haven't heard from the OP since post #1. Let's agree to disagree on this negative area thing.
That's my thought - neither approach does any harm.

We do need to hear from OP ... I was gonna mention it but I noticed that post #1 was not that long ago.

Whatever - one of our approaches is bound to resonate with OPs prior learning.
 
  • #17
Office_Shredder
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The real tragedy is that his confusion might have been that if the graph passes below the x-axis, then the area "under the graph" is all the area literally below the graph. Since there's no x-axis bounding the graph from below, that area appears to be infinite. And then you guys scared him off with your technobabble about negative areas :tongue:
 
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  • #18
Simon Bridge
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The problem is to find the area bounded by the curves ##f(x)=2x+4## and ##g(x)=x^2-4##

OP did the following:
$$\text{Area}=\int_{-2}^4 (2x+4)dx - \int_2^4 (x^2-4)dx + \int_{-2}^2(x^2-4)dx$$

... the error was that the last integral should have been subtracted, not added.
I had hoped that the "negative area" comment would have caused a re-evaluation of the last term... or, at least, a query about what it means. In this case: "negative of the area".

Since g(x) < f(x), OP could have proceeded just by integrating the difference of the functions, and it would have come out automatically... as Mark44 suggested (described by summing the strips).

It may have been more direct just to ask if the contribution from the last integral should be to increase or decrease the overall area ... and follow up with whether the calculation does this.
 
  • #19
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I already got the answer right after I read #2 that day.
Didn't expect so many subsequent replies, thanks everyone.

However, despite my answer being correct, I am confused with the marking scheme's version which did it like this:

∫(-2 to 4) (2x+4) - (xsquare -4) dx

...
..
.
36 unitsquare.

Very elegant, but I can't understand what does it mean.
 
  • #20
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What they are doing here is exactly what I described in post #5. Across the interval [-2, 4], the line y = 2x + 4 is above the parabola y = x2 - 4. If you divide up the interval into subintervals of width Δx, your typical area element will be [(2xi + 4) - (xi2 - 4)]Δx, with xi being some number in the i-th subinterval. If you add up the areas of these typical area elements, you'll get a number close to the actual area.

Better yet, you can write the integral, which will give you the exact area.
$$\int_{-2}^4 2x + 4 - (x^2 - 4)dx$$

The similarity between the formula for the typical area element and the integrand above is not a coincidence.

I wrote the integral using LaTeX. To see what I did, right-click anywhere in the integral.
 
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  • #21
Simon Bridge
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