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Homework Help: Calc help!

  1. Dec 11, 2003 #1
    Ummmm.....

    If h(x)= -h(-x) for all x, what is [tex] \int_{-a}^{a} h(x) \,dx [/tex]

    ????
     
  2. jcsd
  3. Dec 11, 2003 #2
    Draw a picture! It should be clear. What does h(x)=-h(-x) mean? What does a definite integral represent if h(x) is positive?

    Think of h(x)=x3 for instance. This should make the answer clear.
     
  4. Dec 12, 2003 #3

    ShawnD

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    Science Advisor

    It means it's flipped around the origin.

    That question must want a really really generic answer because if it wanted anything specific, you could just make up an answer very easily.
     
  5. Dec 12, 2003 #4
    There's only one possible answer.
    0
    [tex] \int_{-a}^{a} h(x) \,dx =\int_{-a}^{0} h(x)dx + \int_{0}^{a}h(x)dx=\int_{-a}^{0}-h(-x)dx+\int_{0}^{a}h(x)dx[/tex]
    Suppose [tex]H(x)+C=\int h(x) \,dx [/tex]
    [tex]\int_{-a}^{0}-h(-x)dx+\int_{0}^{a}h(x)dx=\int_{0}^{-a}h(-x)dx+\int_{0}^{a}h(x)dx[/tex]
    Using u=-x, du=-dx
    [tex]\int_{0}^{-a}h(-x)dx+\int_{0}^{a}h(x)dx=-\int_{0}^{a}h(u)du+\int_{0}^{a}h(x)dx[/tex]
    [tex]=-H(a)+H(0)+H(a)-H(0)=0[/tex]
     
    Last edited: Dec 12, 2003
  6. Dec 12, 2003 #5

    ShawnD

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    Science Advisor

    What you just wrote is true with EVERY function that has a real domain and is in the form a^b where b is an integer and NOT a variable. Lets look at the integration of X^2 between -1 and 1 then between -1 and 0 then 0 and 1.
    from -1 to 1 = 2/3
    from -1 to 0 and 0 to 1 = 1/3 + 1/3 = 2/3

    Lets try another equation, this time something like (5 - 2x)^4 - 10x +7 between -10 and 10
    from -10 to 10 = 1052640
    from -10 to 0 and 0 to 10 = 976820 + 75820 = 1052640

    That's just way too generic to be the answer he's looking for.
     
    Last edited: Dec 12, 2003
  7. Dec 12, 2003 #6
    f(x)=x^2 does not satisfy the hypothesis f(x)=-f(-x)
    f(x)=x^2 is EVEN
    f(x)=-f(-x) means f is ODD
     
  8. Dec 12, 2003 #7
    Definte Integrals includes + as well as - signs
    area below the x-axis is -
    area above the x-axis is +

    Now the function is odd therefore it issymmetric with I & III quadrant

    Hence if one area is positve the other will be negative.
    result the integral will be zero under the limits i repeat under the limits -a to +a
     
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