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!

Integral Bounds

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data
    This is the problem with the solution:

    upload_2015-2-24_18-40-30.png

    Can someone please explain how the new bounds were computed, I don't quite understand what's going on with the inequalities?

    Also, in the final two steps, how can the f(u) change to f(x)?
     
  2. jcsd
  3. Feb 24, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The inequalities are just another mathematical expression which describes the bounds of integration.

    0 ≤ x ≤ a is just math talk for "all x between 0 and a, including both 0 and a."

    After making the substitution u = x - a, the bounds of the u-integral are obtained by substituting the bounds of the x-integral (0 ≤ x ≤ a) into the expression for u.

    Also, in the final two steps, how can the f(u) change to f(x)?[/QUOTE]

    Once you have the integrand in the same form after u-substitution as the original form of the integrand for the x-integral, you can change u to x or whatever. These are essentially 'dummy' variables at this point.
     
  4. Feb 24, 2015 #3
    Okay, when you say: After making the substitution u = x - a, the bounds of the u-integral are obtained by substituting the bounds of the x-integral (0 ≤ x ≤ a) into the expression for u.

    How would you substitute an inequality (0 ≤ x ≤ a) into an equation (u = x - a) ?

    Thanks for the help.
     

    Attached Files:

  5. Feb 24, 2015 #4

    Mark44

    Staff: Mentor

    I think they have the second set of inequalities backwards; i.e., ##a \le u \le 0##.
    The substitution u = a - x is equivalent to u + x = a. Both u and a range between 0 and a, but as x increases from 0 to a (assumed to positive, or at least nonnegative), u decreases from a to 0.
    The x and u in the two integrals are dummy variables. With regard to integration, ##\int f(u) du## is exactly the same as ##\int f(x) dx##. It's only when you undo the substitution that you get something different.
     
  6. Feb 24, 2015 #5
    Like, when the bounds are numbers, I have no problem changing the bounds when doing the substitution, but here I can't seem to wrap my head around it since the bounds are not 'numbers'.
     
  7. Feb 24, 2015 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You're overthinking this. You don't substitute the inequality into the expression for u, you substitute the values at the ends of the interval into the expression.

    If u = x - a, what's the value of u when x = 0? When x = a?
     
  8. Feb 24, 2015 #7
    When x = 0, the value would be -a, and when x = a, the value would be 0.

    Wow, it was that simple! Hehe.
     
  9. Feb 24, 2015 #8

    Mark44

    Staff: Mentor

    The substitution is u = a - x.
     
  10. Feb 24, 2015 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Look at u = a-x for 0 <= x <= a. When x = 0, what is the value of u? When x = a, what is the value of u?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integral Bounds
  1. Bounded integral (Replies: 2)

  2. Integral bounds (Replies: 2)

Loading...