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Composite function and definite integration

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x)=x^3-\frac{3x^2}{2}+x+\frac{1}{4}[/tex]
    find[tex]\int_{\frac{1}{4}}^{\frac{3}{4}} f(f(x))dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I am clueless here. I started by writing f(f(x)) as
    [tex](f(x))^3-\frac{3(f(x))^2}{2}+f(x)+\frac{1}{4}[/tex]
    I don't think expanding (f(x))^3 would be of any use because this question is from a test paper and it would take a long time if i expanded (f(x))^3 and the other terms. I don't have any idea how to proceed. I observed that the f(x) is always an increasing function, therefore only one real root for f(x) exist but i don't think that would be of any use here.

    Any help is appreciated!
     
  2. jcsd
  3. Nov 3, 2012 #2

    I like Serena

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    Hey Pranav-Arora!

    Is this about a multiple choice perhaps?

    Then you'd be supposed to make an estimate (using inscribed rectangles) and match the answer.
    Or alternatively make a drawing and measure the area.

    I agree that expanding everything is rather time consuming.
     
  4. Nov 3, 2012 #3
    Hey ILS! :smile:

    Nope, this question was from a section where the answer is an integer from 0 to 9.

    I don't understand why are you talking about rectangles here. Maybe i could estimate the area using the graph separately for (f(x))^3 and (f(x))^2.

    Isn't there any proper method to get an exact answer?
     
  5. Nov 4, 2012 #4

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    Ah, so you do have at least some limitation on the acceptable answers.

    An integral is the area under a graph.
    220px-Integral_as_region_under_curve.svg.png

    A first estimate is the area of the rectangle that has a height equal to a point on the graph, and a width equal to the boundaries.
    Somewhat like this (for an integral with boundaries 0.5 and 1.5):
    http://www.livemath.com/hoffnercalculus/preview/Chapter%206/Chpt6_Resources/image37.gif [Broken]

    In your case you could take f(f(1/2)) as the height and (3/4 - 1/4) as the width.
    I believe that will suffice to find the proper answer, since that has to be an integer.

    The method to get the exact answer is exactly what you were already doing.
    I do not see ways to significantly speed up that process.
     
    Last edited by a moderator: May 6, 2017
  6. Nov 4, 2012 #5

    lurflurf

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    Is there a typo? I do not think that integral is an integer.
    Are you familiar with

    [tex](b-a)L\le \int_a^b f(x) dx \le (b-a)U[/tex]
    where U and L are numbers such that
    L<f<U
     
  7. Nov 4, 2012 #6

    SammyS

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    There are several things to notice here.

    The point, (1/2, 1/2), is a fixed point. So, f(1/2) = 1/2, and f(f(1/2)) = 1/2 .

    Also, the point, (1/2, 1/2) is a point of inflection for f(x), and f(x) is a cubic function, so the graph, y = f(x) is invariant under a 180° rotation about the fixed point, (1/2, 1/2). In other words, if you shift the graph, y = f(x), to the left by 1/2 and down by 1/2, the resulting graph, y = f(x + 1/2) - 1/2 is symmetric with respect to the origin.

    Thus, the function, g(x) = f(x + 1/2) - 1/2 is an odd function, actually an odd cubic function.

    The function, g(g(x)), is also an odd function, so [itex]\displaystyle \int_{-1/4}^{1/4}g(g(x))\,dx=0\ .[/itex]

    You can also show, graphically or otherwise, that y=g(g(x)) is a shifted version of y=f(f(x)) .

    So I like Serena's idea of using a rectangle is relevant.
     
  8. Nov 5, 2012 #7

    SammyS

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    Better yet:

    Define [itex]\displaystyle g(x) = f(x+(1/2)) - 1/2[/itex], which is an odd (cubic) function.

    Now for the integral [itex]\displaystyle \int_{1/4}^{3/4}f(f(x))\,dx\ :[/itex]

    Do the substitution x = u + 1/2, then dx = du. x = 1/4, 3/4 becomes u = -1/4, 1/4 .

    [itex]\displaystyle \int_{1/4}^{3/4}f(f(x))\,dx[/itex]
    [itex]\displaystyle =\int_{-1/4}^{1/4}f(f(u + 1/2))\,du[/itex]

    [itex]\displaystyle =\int_{-1/4}^{1/4}f(f(u + 1/2))-\frac{1}{2}+\frac{1}{2}\,du[/itex]

    [itex]\displaystyle =\int_{-1/4}^{1/4}f(f(u + 1/2))-\frac{1}{2}\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du[/itex]

    [itex]\displaystyle =\int_{-1/4}^{1/4}f\left(f(u + 1/2)-1/2 +\frac{1}{2}\right)-\frac{1}{2}\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du[/itex]

    [itex]\displaystyle =\int_{-1/4}^{1/4}f\left(g(u) +\frac{1}{2}\right)-\frac{1}{2}\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du[/itex]

    [itex]\displaystyle =\int_{-1/4}^{1/4}g(g(u))\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du[/itex]
     
  9. Nov 5, 2012 #8

    I like Serena

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    I did not analyze as far as you did, but if I understand you correctly, the area of the rectangle is equal to the exact solution. ;)
    And indeed, that is not a whole number.
     
  10. Nov 5, 2012 #9

    SammyS

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    Yes, that's correct.

    It is the reciprocal of a whole number.
     
  11. Nov 5, 2012 #10
    Woops, sorry for a late reply.

    Sorry, there is a typo. The reciprocal of the definite integral is the answer.
    Yes, i am familiar with that. Thanks for that, this might be useful to estimate the answer. :smile:

    Nicely explained SammyS. :smile:

    Excellent! How do they expect us to think like this and that too in the examination room.
    The solution would be take me some time to understand it completely, i will get back to this thread if i get stuck.

    I will mention it again, there's a typo. The answer is the reciprocal of the value of the definite integral.
     
    Last edited: Nov 5, 2012
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