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Math Help - [SOLVED] Limit of integral question when a substitution is made

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Limit of integral question when a substitution is made

    If I understood well the change of the limits of the integral when we make a substitution, it must looks like this :
    Say we have \int_0^1 3(3x+5)^4dx. Let u=3x+5, then du=3dx. So we have \int_5^8 u^4du=\frac{u^5}{5} \big| _5^8. I changed the limits of the integral of 0 and 1 to 5 and 8 because when x=0, 3x+5=5, etc. But now comes the step I don't understand : \int_5^8 u^4du=\frac{u^5}{5} \big| _5^8=\frac{(3x+5)^5}{5}\big| _0^1. Why the limits change again? It seems totally worthless to bother to change the limits of the integral if at last we change them as they were! I don't understand the why of we change the limits in the last step. (I understand why in the first step, it's because of the chain rule).
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    Moo
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    Hello,

    I don't see why there is this last step either. Maybe it's because they want to show you that both are possible...

    "the chain rule" ?

    ---------------------------
    You can see it this way for both possibilities.

    The first one is what is written.

    The second one, write :

    \int_{x=0}^{x=1} u^4 \ du=\frac{u^5}{5} \bigg|_{x=0}^{x=1}

    and here, replace u.
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    MHF Contributor arbolis's Avatar
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    So there's no deal with changing the limits of the integral.
    "the chain rule" ?
    Yes. It says that if f and g' are continuous, then \int_{g(a)}^{g(b)}f(u)du=\int _a^b f(g(x))\cdot g'(x)dx.
    It gives as an example : \int_a^b \tan (x)dx=\int_a^b \frac{\sin (x)}{\cos (x)}dx=\int_a^b \frac{1}{\cos (x)}\cdot \sin (x)dx =-\int_a^b \frac{1}{\cos (x)}\cdot (-\sin (x))dx. Now let g(x)=\cos (x), g'(x)=-\sin (x), f(u)= \frac{1}{u}.
    \Rightarrow -\int_{\cos (a)}^{\cos (b)} \frac{du}{u}=- \ln|u| \bigg| _{\cos (a)}^{\cos (b)}=\ln \big| \frac{\cos (a)}{\cos (b)} \big|. So afterall changing the limits is obligated? I'm confused now... I'll think about it.
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    Moo
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    Quote Originally Posted by arbolis View Post
    So there's no deal with changing the limits of the integral.
    Yes. It says that if f and g' are continuous, then \int_{g(a)}^{g(b)}f(u)du=\int _a^b f(g(x))\cdot g'(x)dx.
    It gives as an example : \int_a^b \tan (x)dx=\int_a^b \frac{\sin (x)}{\cos (x)}dx=\int_a^b \frac{1}{\cos (x)}\cdot \sin (x)dx =-\int_a^b \frac{1}{\cos (x)}\cdot (-\sin (x))dx. Now let g(x)=\cos (x), g'(x)=-\sin (x), f(u)= \frac{1}{u}.
    \Rightarrow -\int_{\cos (a)}^{\cos (b)} \frac{du}{u}=- \ln|u| \bigg| _{\cos (a)}^{\cos (b)}=\ln \big| \frac{\cos (a)}{\cos (b)} \big|. So afterall changing the limits is obligated? I'm confused now... I'll think about it.
    Ah I see for the "chain rule"

    No, changing the limits is not 'obligated' (compulsory ?) ^^ At least that's what I can conclude by looking at different solutions posted in this forum.. ! They do what they call "backsub" and then apply former limits...
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    MHF Contributor arbolis's Avatar
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    They do what they call "backsub" and then apply former limits...
    I think I know what is a backsub, but not what is to apply former limits. Can you show me a link to such a post, if it's not too much asked.
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    Moo
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    Quote Originally Posted by arbolis View Post
    I think I know what is a backsub, but not what is to apply former limits. Can you show me a link to such a post, if it's not too much asked.
    By formal limits, I mean a and b. Sorry for not being clear

    It corresponds to this :

    <br />
\int_{x=0}^{x=1} u^4 \ du=\frac{u^5}{5} \bigg|_{x=0}^{x=1}<br />

    and here, replace u
    that's what I call "backsub and then apply former limits"

    ------
    En gros, tu retransformes u en fonction de x et tu appliques à ton résultat les premières limites de l'intégrale, c'est-à-dire a et b dans ton post #3



    But do it as you wish to. Either you change the limits, either you backsubstitute
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    By formal limits, I mean a and b. Sorry for not being clear

    It corresponds to this :


    that's what I call "backsub and then apply former limits"

    ------
    En gros, tu retransformes u en fonction de x et tu appliques à ton résultat les premières limites de l'intégrale, c'est-à-dire a et b dans ton post #3



    But do it as you wish to. Either you change the limits, either you backsubstitute
    A backsub, is, or at least is how I use it, when you complete a limit you cannot leave it in terms of u's since the original function was a function of x. So you must backsub what u was defined as in terms of x.
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    MHF Contributor arbolis's Avatar
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    et tu appliques à ton résultat les premières limites de l'intégrale, c'est-à-dire a et b dans ton post #3
    In my post 3 I do a backsub but I don't change the limits to obtain the first limits. I had the limits a and b and I finish with \cos (a) and \cos (b), I don't understand why.
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    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by arbolis View Post
    Why the limits change again? It seems totally worthless to bother to change the limits of the integral if at last we change them as they were! I don't understand the why of we change the limits in the last step.
    A reason might be that, when computing an anti-derivative instead of a definite integral, the answer has to be, in this case, in terms of x and not in terms of u : one should write \int3(3x+5)^4\,\mathrm{d}x=\frac{(3x+5)^5}{5}+C and not \int3(3x+5)^4\,\mathrm{d}x=\frac{u^5}{5}+C. Substituting u=3x+5 at the end of the computation might be a means to avoid mistakes : if you always do this (i.e. even for integrals although it is useless) your answer will always be right and you won't think about giving the expression of the anti-derivatives in terms of u when it's not allowed.
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    MHF Contributor arbolis's Avatar
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    A reason might be that, when computing an anti-derivative instead of a definite integral, the answer has to be, in this case, in terms of and not in terms of : one should write and not . Substituting at the end of the computation might be a means to avoid mistakes : if you always do this (i.e. even for integrals although it is useless) your answer will always be right and you won't think about giving the expression of the anti-derivatives in terms of when it's not allowed.
    Hmm, I don't understand very well. I know that the solution must be in term of x and not u. For an indefinite integral it's much more simple to find the anti-derivatives : we don't have limits. While when we are working on a definite integral, it seems to me that sometimes the integral is equal to an anti-derivative evaluated in the same limits of the integral and sometimes no.
    For example, same limits : \int_0^1 3(3x+5)^4dx=\frac{(3x+5)^5}{5}\big| _0^1+C. Different limits : \int_a^b \tan (x)dx=- \ln|u| \bigg| _{\cos (a)}^{\cos (b)}+C. Hmm wait! Here if we replace u by \cos (x) and we change the limits of - \ln|u| \bigg| _{\cos (a)}^{\cos (b)} into a and b, I think it works... so ok! When I do a backsub, I always change the limits to get the first limits! Lol, seems useless to bother the limits then... I could just solve a definite integral as if it was an indefinite one and at the end I put the same limits and all work.
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I could just solve a definite integral as if it was an indefinite one and at the end I put the same limits and all work.
    This is almost always what I do.
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