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Integral question

  1. Jan 16, 2008 #1
    x^2 * [1+(sinx)^2007]dx

    the integral is from 1 to -1

    i tried to solve it by parts but its too complicated
     
  2. jcsd
  3. Jan 16, 2008 #2

    Tom Mattson

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    Have you learned anything about integrating even and odd functions?
     
  4. Jan 16, 2008 #3
    Can you tell me a little more? Thanks :-]
     
  5. Jan 16, 2008 #4
    Even functions satisfy f(-x) = f(x). Odd functions satisfy f(-x) = -f(x). Note the interval over which you are integrating makes this rather useful.
     
  6. Jan 17, 2008 #5
    thats not an odd functon my function is composed of two different function
    which are different from one another
    f(-x) = -f(x)

    cosx is an odd function ,
    but here i am not sure
    if i knew its odd then we need to split it into two different intervals
    and still find the integral

    i tried to solve it by parts but thats not working
     
  7. Jan 17, 2008 #6
    Indeed - but it is a linear combination of two functions - one even and one odd.
     
  8. Jan 17, 2008 #7

    Gib Z

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    Think of the integrand as one entire function, not the composition of separate ones. Check if its even or odd. If you haven't run into integrating odd and even functions before, you can interpret it algebraically or geometrically.

    Algebraically, split the integral over the intervals -1 to 0 and 0 to 1. Then use the odd function property.

    Geometrically, do you see any symmetry when you graph the functions?
     
  9. Jan 17, 2008 #8

    HallsofIvy

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    By symmetry, if f(x) is an even function,
    [tex]\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx[/tex]
    If f(x) is an odd function,
    [tex]\int_{-a}^a f(x)dx= 0[/tex]

    Write your integrand as x2+ x2 sin2007(x).

    What is (-x)2? What is (-x)2 sin2007(-x)?
     
  10. Jan 17, 2008 #9
    my problem is to solve the actual function to find the previos function
    on which we made the derivative

    by parts is not working
     
  11. Jan 17, 2008 #10
    Read HallsofIvy's post.

    Expand the product out, and take the integral of each of the terms, using the idea of odd and even functions.

    Why would u use by-parts in the first place? What are you trying to reduce to zero power? If you're trying to eventually take x^2 down to x^0, I don't see how you can in turn, integrate sin(x)^2007.
     
  12. Jan 17, 2008 #11
    i read the post and in the end i am still required to solve
    x^2 * (sinx)^2007

    and i dont have any clue how to do it
     
  13. Jan 17, 2008 #12

    Gib Z

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    [tex]f(x) = x^2 \cdot \sin^{2007} (x)[/tex]

    [tex]f(-x) = (-x)^2 \cdot \sin^{2007} (-x) = x^2 \cdot \left( - \sin^{2007} x\right)= - x^2 \cdot \sin^{2007} (x) = - f(x)[/tex]

    Repeating;
    [tex]f(-x) = - f(x)[/tex].

    [tex]\int^a_{-a} f(x) dx = \int^0_{-a} f(x) dx + \int^a_0 f(x) dx[/tex].

    In the first integral, letting x = -u, dx = - du. The bounds change from 0 to 0, and -a to a.

    [tex]\int^0_{a} - f(-u) du = \int^0_a f(x) dx = - \int_0^a f(x) dx[/tex].

    I probably shouldn't have given you the whole answer, but I trust that you will read over the solution several times so you understand it perfectly, or else I just did a misdeed.
     
  14. Jan 17, 2008 #13
    but you didnt finish it
    the main problem is to solve f(x)dx
    from a to 0
    or the other one

    iknow of the proccess of spliting the interval
    the main problem for me is to solve this integral f(x)dx

    i tried to solve it in part ,in substitution
    nothing works
     
  15. Jan 17, 2008 #14

    Gib Z

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    Do you just want an anti derivative or do you want the smart way???

    I practically did it for you!!

    [tex]\int^a_{-a} f(x) = \int^a_0 f(x) dx - \int^a_0 f(x) dx[/tex].What is the right hand side equal to?
     
  16. Jan 17, 2008 #15
    it equals to zero
    wow
    you switched the intervals
     
  17. Jan 17, 2008 #16

    Gib Z

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    =] So you fine with this now? We can go over other examples if you need to.
     
  18. Jan 17, 2008 #17
  19. Jan 17, 2008 #18

    Defennder

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    I'm wondering, does anyone know how to evaluate the function as an indefinite integral? Any hints?
     
  20. Jan 17, 2008 #19

    Gib Z

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    The anti derivative will be ugly, if it exists in elementary form.
     
  21. Jan 23, 2008 #20
    why its zero???


    because if we draw the function we get two areas which are
    opposite one to another

    bu still these areas exists and we need to count them
    they both equal to each other
    so we need to find the area of one of them
    and multiply the result by 2

    ???

    why its zero
    zero is a mistake
    we need to find the area of one of them
    and multiply the result by 2
     
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