1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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

    User Avatar
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calc help!
  1. Calc Help! (Replies: 1)

  2. Help with calc (Replies: 5)

  3. Calc Help (Replies: 1)

  4. Calc Help (Replies: 1)

Loading...