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Definite integration, even function. confused about proof

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f[/itex] be integrable on the closed interval [itex][-a,a][/itex]

    If [itex]f[/itex] is an even function, then
    [itex]\int^a_{-a}f(x)\,dx[/itex] = [itex]2\int^a_0f(x)\,dx[/itex]

    Prove this.


    2. Relevant equations



    3. The attempt at a solution
    The solution is given in the book.

    Because [itex]f[/itex] is even, you know that [itex]f(x) = f(-x)[/itex]. Using the substitution [itex]u = -x[/itex] produces

    [itex]\int^0_{-a}f(x)\,dx =\int^0_af(-u)\,(-du) = -\int^0_af(u)\,du
    =\int^a_0f(u)\,du = \int^a_0f(x)\,dx[/itex]

    Now that part that confuses me is [itex]\int^a_0f(u)\,du = \int^a_0f(x)\,dx[/itex]

    Wouldnt [itex]u=-x[/itex] mean [itex]du=-dx[/itex] which would produce [itex]\int^a_0f(u)\,du = -\int^a_0f(x)\,dx[/itex]

    I know my reasoning must fall apart somewhere, since that would mean

    If [itex]f[/itex] is an even function, then [itex]\int^a_{-a}f(x)\,dx = 0[/itex].

    I just cannot see how my reasoning is wrong.

    If it makes any difference this is the remainder of the proof in the book.

    [itex]\int^a_{-a}f(x)\,dx =\int^0_{-a}f(x)\,dx + \int^a_0f(x)\,dx
    =\int^a_0f(x)\,dx + \int^a_0f(x)\,dx = 2\int^a_0f(x)\,dx[/itex]
     
  2. jcsd
  3. Apr 13, 2013 #2
    You can label the integration variable in a definite integral with any letter, it does not change anything. The definite integral is completely determined by the function under the integral sign and its limits. ## \int_a^b f(x)dx = \int_a^b f(u)du = \int_a^b f(\xi)d\xi = \int_a^b f(\mathfrak{S})d\mathfrak{S} ##
     
  4. Apr 13, 2013 #3
    No because switching [itex] u \leftrightarrow x [/itex] you also have to switch the limits of
    integration, which is what you did in the first place but don't do here!

    [itex]u=-x[/itex] mean [itex]du=-dx[/itex] which would produce [itex]\int^a_0f(u)\,du = -\int^{-a}_0 f(x)\,dx[/itex]
     
  5. Apr 14, 2013 #4
    Oh right. I see the mistake I made. Thanks qbert.

    But I still dont see how that leads to [itex]\int^0_{-a}f(x)\,dx = \int^a_0f(x)\,dx[/itex]

    I was looing at it as if the author was using a theorem he proved earlier in the text:

    If [itex]u=g(x)[/itex] has a continuous derivative on the closed interval [itex][a,b][/itex] and [itex]f[/itex] is continuous on the range of [itex]g[/itex], then

    [itex]\int^b_af(g(x))g'(x)\,dx = \int^{g(b)}_{g(a)}f(u)\,du[/itex]

    Im not completley sure but, it would seem the author did use it until he got to the point [itex]\int^a_0f(u)\,du = \int^a_0f(x)\,dx[/itex]

    Are you saying that when the author got to that point he didnt change the variable, but just relabled it?
     
  6. Apr 14, 2013 #5
    A real change of variables typically changes both the integrand and the limits. This is what the author had till the final step. Then, yes, he simply re-labeled the variable. How can you tell? It did not change the integrand, nor the limits.
     
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