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Math Help - unit of integration vs unit of boundaries

  1. #1
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    unit of integration vs unit of boundaries

    Hi
    Was just wondering:
    In the expression
    \int_{a}^{b}f(x)dx
    it is assumed that a and b are values of x.
    Yet when we do u-substitution, to get
    \int_{a}^{b}g(u)du
    we are still assuming that a and b are value of x (the boundaries of the area we are finding, as it were).
    But really, shouldn't they also be converted to the corresponding values of u? The unit of integration is now u and everything else in the expression is now in terms of u, yet these boundaries are still referring to values of x. To change f(x) -> g(u) we needed some function mapping x to u, so we must also be able to map the boundaries to u as well.

    The conventional method when using u-ubstitution to help with integration involves re-converting the final expression (e.g. 4u^2 +2u +c) back to x, and then substituting in the boundary values of x.
    If the boundary values are converted from x-values to u-values, this reversal wouldn't be required before the final evaluation.

    While I admit that it is probably just easier to do it the conventional way, my question is whether or not the procedure outlined above is actually correct.

    Cheers
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chug1 View Post
    Hi
    Was just wondering:
    In the expression
    \int_{a}^{b}f(x)dx
    it is assumed that a and b are values of x.
    Yet when we do u-substitution, to get
    \int_{a}^{b}g(u)du
    we are still assuming that a and b are value of x (the boundaries of the area we are finding, as it were).
    But really, shouldn't they also be converted to the corresponding values of u? The unit of integration is now u and everything else in the expression is now in terms of u, yet these boundaries are still referring to values of x. To change f(x) -> g(u) we needed some function mapping x to u, so we must also be able to map the boundaries to u as well.

    The conventional method when using u-ubstitution to help with integration involves re-converting the final expression (e.g. 4u^2 +2u +c) back to x, and then substituting in the boundary values of x.
    If the boundary values are converted from x-values to u-values, this reversal wouldn't be required before the final evaluation.

    While I admit that it is probably just easier to do it the conventional way, my question is whether or not the procedure outlined above is actually correct.

    Cheers
    You're ideology is correct.

    You can actually convert the boundaries of x to boundaries of u!

    If u=h\!\left(x\right) is your substitution, then your new limits of integration (i.e. the boundaries) are a_u=h\!\left(a\right) and b_u=h\!\left(b\right).

    So, \int_a^bf\!\left(x\right)\,dx\xrightarrow{u=h\!\le  ft(x\right)}{}\int_{a_u}^{b_u}g\!\left(u\right)\,d  u.

    Once this is done, you don't have to convert back to x!
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  3. #3
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    Quote Originally Posted by chug1 View Post
    Hi
    Was just wondering:
    In the expression
    \int_{a}^{b}f(x)dx
    it is assumed that a and b are values of x.
    Yet when we do u-substitution, to get
    \int_{a}^{b}g(u)du
    we are still assuming that a and b are value of x (the boundaries of the area we are finding, as it were).
    But really, shouldn't they also be converted to the corresponding values of u? The unit of integration is now u and everything else in the expression is now in terms of u, yet these boundaries are still referring to values of x. To change f(x) -> g(u) we needed some function mapping x to u, so we must also be able to map the boundaries to u as well.

    The conventional method when using u-ubstitution to help with integration involves re-converting the final expression (e.g. 4u^2 +2u +c) back to x, and then substituting in the boundary values of x.
    If the boundary values are converted from x-values to u-values, this reversal wouldn't be required before the final evaluation.

    While I admit that it is probably just easier to do it the conventional way, my question is whether or not the procedure outlined above is actually correct.

    Cheers
    Yes, basically you're thinking of the change of variables theorem that says: If you have f:[a,b] \rightarrow \mathbb{R} and a function g:[c,d] \rightarrow [a,b] that is continously differentiable and with a continously diff. inverse then \int_{g(c)} ^{g(d)} \ f(x)dx = \int_{c} ^{d} \ f(g(u))g'(u)du
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  4. #4
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    Thank you both very much!
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