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Homework Help: Calculating integrals of area between two functions, involving absolute values

  1. Jan 3, 2009 #1
    Hi all,

    I'm currently preparing for pre-tertiary mathematics, studying from Apostol's "One-Variable Calculus". I have just begun to work on the theory of integration of trigonometric functions, but I found that with the last set of exercises (on finding area between two functions, over some interval) I wasn't able to complete them (the issue was with absolute value functions).

    Any assistance would be greatly appreciated.

    1. The problem statement, all variables and given/known data

    Take one, for example, with these functions (over the interval [-1, 1]):

    [tex] f(x) = |x| [/tex]
    [tex] g(x) = x^2-1 [/tex]

    I recognise that over this interval, [tex] f(x) > g(x) for all x [/tex], so the answer should be equal to the integral of the difference f(x) - g(x) with respect to x over the previously said interval.

    How do I tackle this from here? My graphics calculator software tells me the area is equal to 2.33..., but I'm not sure how to arrive at this value.

    2. Relevant equations

    [tex]

    \int_{a}^{b} [f(x) - g(x)] dx

    [/tex]

    3. The attempt at a solution

    To show that I can work these integrals, I first show another example which may assist to solve the original example.

    [tex] f(x) = x(x^2 - 1) [/tex]
    [tex] g(x) = x [/tex]

    The interval we must evaluate this integral on is as follows: [tex] [-1, 2^{1/2}] [/tex].

    Over the interval [0, 1], f(x) > g(x). Over the interval [tex] [0, 2^{1/2}] [/tex], g(x) > f(x).

    So to formalise this now, we have the following:


    [tex]

    \int_{-1}^{0} [x(x^2 - 1) - x] dx + \int_{0}^{2^{1/2}} [(x - x^3 - x)] dx [/tex]

    [tex] = -\int_{0}^{-1} [x^3 - x - x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
    [tex] = -\int_{0}^{-1} [x^3 - 2x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
    [tex] = -\left[ \frac{x^4}{4} - {x^2} \right]_{0}^{-1} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
    [tex] = [-\frac{-1^4}{4} + -1^2] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
    [tex] = [-\frac{1}{4} + 1] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
    [tex] = \frac{3}{4} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
    [tex] = \frac{3}{4} + \left[ \frac{-x^4}{4} + {x^2} \right]_{0}^{2^{1/2}} [/tex]
    [tex] = \frac{3}{4} + [-\frac {(2^{1/2})^4}{4} + (2^{1/2})^2] = \frac{3}{4} +[-\frac {4}{4} + 2] [/tex]
    [tex] = \frac{3}{4} - 1 + 2 = \frac{3}{4} + 1 [/tex]
    [tex] = \frac{3+4}{4} = \frac{7}{4} = 1.75. [/tex]

    Now, I believe that, perhaps with the absolute value example, you can substitute values in, similarly to this example I've worked. First, I note that both these functions are monotonically decreasing over the interval [-1, 0] and increasing over [0, 1].

    So, I'll try:

    [tex] \int_{-1}^{0} [f(x) - g(x)] dx + \int_{0}^{1} [f(x) - g(x)] dx [/tex]
    [tex] = -\int_{0}^{-1} [|x| - (x^2 - 1)] dx + \int_{0}^{1} [|x| - (x^2 - 1)] dx [/tex]
    [tex] = \left[ \frac{|x^2|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{-1} + \left[ \frac{|x^2|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{1} [/tex]
    [tex] = 0 - [\frac{1}{2} - (\frac{-1^3}{3} +1)] + [\frac{1}{2} - (\frac{1}{3} - 1)] [/tex]
    [tex] = [-\frac{1}{2} + (-\frac{1^3}{3} + 1)] + [\frac{1}{2} - \frac{1}{3} + 1] [/tex]
    [tex] = - \frac{1}{2} - \frac{1}{3} + 1 + \frac{1}{2} - \frac{1}{3} + 1 = \frac{4}{3} [/tex]

    Clearly, I'm out by 1. Where have I gone wrong? Any assistance would be simply grand. I tried with the first integral being positive, but I ended with the same result mysteriously enough. I just don't understand!

    Cheers,
    Davin
     
  2. jcsd
  3. Jan 3, 2009 #2
    Hey,
    You have to drop your absolute value bars on the second line of your solution up there. The equation of the line from -1 to 0 is -x and from 0 to 1 it's just x...that changes the sign on your 1/2 aand gets you what you know is correct.
    CC
     
  4. Jan 3, 2009 #3
    g(x) and f(x) both have symmetry on the y-axis. You can do it like:

    [tex] 2\int_{0}^{1} - x^2 + x - 1 \delta x = 2\left[1 \frac{1}{6} \right][/tex] which is 2.33

    [tex]\int_{-1}^{1} |x| - (x^2 - 1) \delta x = \left[\frac{|x^2|}{2} - \frac{x^3}{3} + x\right]_{-1}^{1} [/tex]

    [tex] = \left[ \frac{1}{2} - \frac{1}{3} + 1\right] - \left[ -\frac{1}{2} + \frac{1}{3} - 1\right] = 2\frac{1}{3}[/tex]

    I think that's right.
     
    Last edited: Jan 3, 2009
  5. Jan 3, 2009 #4
    Absolute Value integrals??

    Thank you very much, guys. Nice to hear that you can solve integrals by looking for symmetry! This I wasn't aware of.

    Gregg, I tried your first integral and I didn't get the same answer as you. Your second integral I understand perfectly though.

    Looking closer, it seems you wrote the first integral out incorrectly. I believe it should be this:

    [tex]
    2 \int_{0}^{1} |x| - (x^2 - 1) \delta x = 2\left[\frac{|x^2|}{2} - \frac{x^3}{3} + x\right]_{0}^{1}
    [/tex]
    [tex] = 2\left[\frac{1}{2} - \frac{1}{3} + 1\right] [/tex]
    [tex] = 2\left[1 \frac{1}{6} \right] = 2\frac{1}{3} [/tex]

    I truly do appreciate your assistance. This ability to solve using symmetry is quite remarkable, but I suppose it really isn't much of a surprise: symmetry is useful in so many areas in mathematics!

    Cheers,
    Davin
     
  6. Jan 3, 2009 #5
    Re: Absolute Value integrals??

    Now I look at it, it can be solved even easier than that. Take the function y = |x|. You know that the area underneath it between -1 and 1 will be 1.

    Take the area under the x-axis for [tex]x^2 - 1 = (x+1)(x-1)[/tex] gives

    [tex]\int_{-1}^{1} x^2 - 1 \delta x = \left[ \frac{x^3}{3} - x \right]_{-1}^{1}[/tex]


    [tex]\int_{-1}^{1} x^2 - 1 \delta x = \left[ \frac{1^3}{3} - 1 \right]-\left[ \frac{-1^3}{3} + 1 \right][/tex]

    [tex]\int_{-1}^{1} x^2 - 1 \delta x = \left[ -\frac{2}{3} \right] - \left[ \frac{2}{3} \right] = - \frac{4}{3}[/tex]

    [tex]1 + \frac{4}{3} = 2 \frac{1}{3}[/tex]

    The reason I find this interesting is because you first answers are out by 1 square unit.

    I think for this reason it safest in this situation to use the method above.

    int.jpg

    Int2.jpg
    The blue curve is [tex]|x| - (x^2-1)[/tex] and the red curve is [tex] x - (x^2-1) [/tex] (they overlap).
    Because x is modulus it has a vertex with the x component of [tex] \pm \frac{1}{2} [/tex].

    I also find that
    [tex] \int_{-1}^{1} -x^2+x+1 \delta x = \frac{4}{3} [/tex]

    I believe this is the reason your answers were out by 1 square unit. The modulus of x was not correctly integrated. But honestly I don't know.
     
    Last edited: Jan 4, 2009
  7. Jan 3, 2009 #6
    Thanks again, Gregg. I personally feel that this case of looking for symmetries is simpler. Then it's a more intuitive approach. But thanks for elaborating further on these ideas. :wink:

    Cheers,
    Davin
     
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