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If f is even then then left and right integrals are equal

  1. Jun 4, 2015 #1
    1. The problem statement, all variables and given/known data
    If ##f## is an even function then $$\int_{-a}^{0} f = \int_{0}^{a} f$$

    2. Relevant equations


    3. The attempt at a solution
    My attempt was trying to show the upper sum of both integrals were equal.

    Take a partition of [-a,0] call it ##P_{1}##, and ##P_{2}## for [0.a].

    if we can show that ##U(f,P_{1}) = U(f,P_{2})## I think that might do it. However, i am not sure how to get there
     
  2. jcsd
  3. Jun 4, 2015 #2
    I think it would be helpful to know that f is even if f(x) = f(-x)
     
  4. Jun 5, 2015 #3
    Just do a substitution, say, y=-x. Then the result follows.
     
  5. Jun 5, 2015 #4
    is that a substitution on the limits or the variable?
     
  6. Jun 5, 2015 #5

    Mark44

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    Yes (to both). If you change the variable, the limits of integration change as well.
     
  7. Jun 5, 2015 #6
    ok i keep getting a negative in there and I can't figure out how to get rid of it. here is my working

    $$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(-x) dx $$

    let ##y = -x## ##\rightarrow## ##dy = -dx##. So we now have ##-\int_{0}^{-a} f(y) dy = \int_{-a}^{0} f(y) dy##

    now if i sub -x back in for y I get a negative infront again since du = -dx. Any thoughts?
     
  8. Jun 5, 2015 #7

    Mark44

    Staff: Mentor

    Yes. When you reverse the order of the integration limits, the sign of the integral changes.

    IOW, ##\int_a^b f(x)dx = -\int_b^a f(x)dx##
     
  9. Jun 5, 2015 #8
    Please do not start with what you need to prove.
    Try to start from LHS and end with RHS.
     
  10. Jun 5, 2015 #9

    SammyS

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    He (or she) is not starting with what needs to be proved.
     
  11. Jun 6, 2015 #10
    Let f be any odd function, then lim x -> -a (f(-a)-f(x))/(-a-x) = (-f(a) - f(x))/(-a-x) = (f(a) + f(x))/(a+x) = (f(a) - f(-x))/(a-(-x)) = lim y -> a (f(a) - f(y))-(a-y)this proves that f'(-a) = f'(a), now since your function is even, it must be a derivative of and odd function , Good luck ;)
    Edit, y = -x , just for the fancy notation !
     
  12. Jun 6, 2015 #11
    Yeah but isn't that not what we want. Maybe i am not seeing something but it seems we wouldnt want the negative out in front?
     
  13. Jun 6, 2015 #12

    vela

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    Try doing the substitution only on one side of the equation you started with.
     
  14. Jun 6, 2015 #13
    lim y -> a (f(a) - f(y))-(a-y) this is meant to be lim y-> a (f(a)-f(y))/(a-y)
     
  15. Jun 6, 2015 #14
    but i have been working on just one side. i do not see a point where i stop working on the side that I "started" with
     
  16. Jun 6, 2015 #15

    vela

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    Oh, I misunderstood what you wrote. You got
    $$\int_0^a f(x)\,dx = \int_0^a f(-x)\,dx = -\int_0^{-a} f(y)\,dy = \int_{-a}^0 f(y)\,dy.$$ Do you see that you've shown what you setting out to prove?
     
    Last edited: Jun 7, 2015
  17. Jun 7, 2015 #16
    But the variables are different no? When you sub -x back in for y wouldnt you get $$-\int_{-a}^{0} f(-x) dx?$$
     
  18. Jun 7, 2015 #17

    vela

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    Remember the variable of integration is a dummy variable.
     
  19. Jun 7, 2015 #18
    I am sorry if I am being difficult vela, but I just dont understand why showing

    $$\int_{0}^{a}f(x)dx = \int_{-a}^{0} f(y) dy$$
    shows that
    $$\int_{0}^{a}f(x)dx = \int_{-a}^{0} f(x) dx$$

    if anything i think i have shown...

    $$\int_{0}^{a} f(x) dx = -\int_{-a}^{0} f(x) dx$$
     
    Last edited: Jun 7, 2015
  20. Jun 7, 2015 #19

    SammyS

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    You may want to use the Edit feature to fix your typo !
     
  21. Jun 7, 2015 #20
    No thats what i meant...

    Since dy = -dx we get that negative out in front done we?
     
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