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Substitution Question

  1. Mar 13, 2008 #1
    http://www.math.cmu.edu/~handron/21_122/hw/hw1.pdf

    My question is about number 5.5.64 on that page. "If f is continuous on R, prove that ..."

    I have been unable to do this and would appreciate help.
     
  2. jcsd
  3. Mar 13, 2008 #2
    Any thoughts so far??
    Try to use the def. of the definite integral, and see if you can come up with sth.
     
  4. Mar 13, 2008 #3
    Hmm no sorry I still dont see it. Are you saying I have to use Riemann sums?
     
  5. Mar 13, 2008 #4
    Just by looking at the integrals, look at how the f(...) changes, that's a large hint to the substitution to use.
    I think they should have the RHS with f(u)du it's a bit confusing with f(x) dx.
     
  6. Mar 13, 2008 #5
    Oh I see, thanks alot. Yeah, the right hand side had me confused.
     
  7. Mar 13, 2008 #6
    hey I just saw a few of these exercises while reviewing substitution/by parts this week. pretty cool how simple geometric arguments can be shown by u-substitution.

    may I ask if are you attending CMU lamoid?
     
  8. Mar 13, 2008 #7
    I'm not at Carnegie Mellon but a close friend of mine is. He's the one who sent me the site. I think he is taking computer science.
     
  9. Mar 13, 2008 #8
    I know I did ask the question, but could someone explain a bit more. I never really got this concept when I actually "learned" it.

    We substitute x+a for u and so du=dx

    But then when we integrate both sides do we just assume there is some function F(u) such that F'(u) = f(u)?
     
  10. Mar 14, 2008 #9

    Gib Z

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    When do we integrate both sides? After letting u=x+a we get [tex]\int^{b+c}_{a+c} f(u) du[/tex] and then all we do is realise the dummy variable can be written as any letter we like, so change it back to x.

    However, for your information, 1) We don't need to assume, you define a function as such. A function defined as the solution to a differential equation is not that extraordinary. 2) There is a theorem which states that if f(x) is continuous, [tex]F(x) = \int^x_a f(t) dt[/tex] is also a continuous function of x.
     
  11. Mar 14, 2008 #10
    We'd need to integrate to apply the limits of integration wouldn't we to show they are indeed equal?

    Also why would you make the substitution u=x on the RHS after you all ready made the substitution u=x+a on the LHS?
     
  12. Mar 14, 2008 #11

    Gib Z

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    I don't understand your first question :(

    You don't NEED to change the variable back to x, but its just a dummy variable, it doesn't affect the answer, so we can. You DO see how [tex] \int^b_a f(t) dt = \int^b_a f(x) dx = \int^b_a f(\phi) d\phi ......[/tex] ?
     
  13. Mar 15, 2008 #12
    To show that:

    [tex]\int^b_a f(x+c) dx = \int^{b+c}_{a+c} f(x) dx[/tex]

    we need to find the anti-derivatives and apply the limits of integration so show that the two are indeed equal, right?

    I know it's just a dummy variable, but the entire expression x+c isn't completely a dummy variable, because if it was then the two equations wouldn't be equal, would they since you have different limits of integration.

    I was not kidding when I said I really never got this concept -.-
     
  14. Mar 15, 2008 #13

    Gib Z

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    Don't worry lol, PF is here to teach =]

    Ok first of all, substitution is basically just replacing a term with nicer looking terms, so that its easier to see how to continue. [itex]( x+ 2\pi + xy)^5 + 2( x+ 2\pi + xy)^4 - 6( x+ 2\pi + xy)[/itex] might seem a bit daunting, but looks much nicer when its [itex]u^5 + 2u^4 -6u[/itex], with the obvious substitution. It's sort of the same for integration.

    The shorthand
    [tex]\int^b_a f(x+c) dx[/tex] really means [tex]\int^{x=b}_{x=a} f(x+c) dx[/tex].

    So when we make the substitution u= x+c, we have to make sure every piece of information becomes in terms of u. [itex] u^5+ 2u^4 -6( x+ 2\pi + xy)[/itex] is quite useless, its only when all parts are in terms of u are things easier.

    So first we take care of the bounds. x=b, so u=b+c. x=a, so u=a+c. f(x+c) = f(u). du=dx.

    Replacing all the parts of the integral with those, it becomes [tex] \int^{u= b+c}_{u=a+c} f(u) du[/tex], which in shorter notation, becomes [tex]\int^{b+c}_{a+c} f(u) du[/tex]

    All this without finding anti-derivatives. You should probably know that most of the theory of Integration can be done without finding anti-derivatives, its just a helpful tool that we can evaluate them with.
     
  15. Mar 15, 2008 #14

    HallsofIvy

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    No, you don't need to find the anti-derivatives- in fact, you can't because you don't know what "f" is.

    It looks to me like a straight forward substitution. Let u= x+ c. Of course, since both "x" and "u are, as Gib Z said, dummy variables (the result is a constant- the same constant whether you use "x", or "u", or any symbol) (no one said that "x+ c" was a dummy variable) then it should be clear that
    [tex]\int_{a+c}^{b+c} f(u)du= \int_{a+c}^{b+c}f(x)dx[/tex].

    ( Once again, Gib Z typed faster than I did!)
     
  16. Mar 15, 2008 #15
    Ah ok, I think I get it now, thanks :)
     
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