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Math Help - Integral substitution

  1. #1
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    Question Integral substitution

    Hello,

    I have an exercise with solution and I don't understand it.

    Calculate:

     \int x \sqrt{x+1}^{3} \mathrm{d} x

    Solution:

    Substitution:  u = g(x) =  \sqrt{x+1}

     \frac{\mathrm{d} u}{\mathrm{d} x} = g'(x) = \frac{1}{2 \sqrt{x+1}} = \frac{1}{2u}

    \sqrt{x+1}^{3} will be substituted by u^{3} and \mathrm{d}x by \dfrac{\mathrm{d}u}{\frac{1}{2u}} = 2u \mathrm{d} u

    Up to here I understand it but now comes the problem.

    The book says, that in the integral is still a x. Okay. Now they solve the equation x+1 = u^{2} and get x = u^{2} -1, so that the hole substitution leads to:

     \int x \sqrt{x+1}^{3} \mathrm{d} x   =   \int (u^{2} - 1) u^{3} ~ 2u \mathrm{d} u

    What are the mathematical reasons for the equation x+1 = u^{2} ?

    I have no idea from where this come from.

    They got it from g(x) but why is this allowed?

    Thanks for help
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Besserwisser View Post
    Hello,

    I have an exercise with solution and I don't understand it.

    Calculate:

     \int x \sqrt{x+1}^{3} \mathrm{d} x

    Solution:

    Substitution:  u = g(x) = \sqrt{x+1}

     \frac{\mathrm{d} u}{\mathrm{d} x} = g'(x) = \frac{1}{2 \sqrt{x+1}} = \frac{1}{2u}

    \sqrt{x+1}^{3} will be substituted by u^{3} and \mathrm{d}x by \dfrac{\mathrm{d}u}{\frac{1}{2u}} = 2u \mathrm{d} u

    Up to here I understand it but now comes the problem.

    The book says, that in the integral is still a x. Okay. Now they solve the equation x+1 = u^{2} and get x = u^{2} -1, so that the hole substitution leads to:

     \int x \sqrt{x+1}^{3} \mathrm{d} x = \int (u^{2} - 1) u^{3} ~ 2u \mathrm{d} u

    What are the mathematical reasons for the equation x+1 = u^{2} ?

    I have no idea from where this come from.

    They got it from g(x) but why is this allowed?

    Thanks for help
    If u=\sqrt{x+1}, then u^2=x+1 so that x=u^2-1

    See?
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  3. #3
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    Hi,

    thats not my problem. I want to know why am I allowed to do this. They solve the function u, thats easy but why?

    Up to this step the integral is

     \int x \sqrt{x+1}^{3} \mathrm{d} x   =   \int x u^{3} ~ 2u \mathrm{d} u

    I don't understand the reason, the mathematical rules what u has to do with the other x in the integral, because the substitution is only u = \sqrt{x+1}.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Did you not say this?
    Quote Originally Posted by Besserwisser View Post
    What are the mathematical reasons for the equation x+1 = u^{2} ?

    I have no idea from where this come from.

    But, anyway....

    If we let u=\sqrt{x+1}, then x=u^2-1 which implies that dx=2udu.

    So, now just make the neccessary substitutions.
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  5. #5
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    I said that I got this calculation to that point. I don't understand what u has to do with the other x in the integral. Why can I calculate this eauation? Not how. The calculation is trivial but I didn't understand the relation between the function u and the other x.

    I guess I didn't see the forrest because of the trees...
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Besserwisser View Post
    The calculation is trivial but I didn't understand the relation between the function u and the other x.

    I guess I didn't see the forrest because of the trees...
    I don't understand what you are asking. Is your problem with differentiation, substitution, or what? Be specific. We have all been where you are, but in order for us to help you, you are going to have to be precise in what you are asking. Mathematics is all about presicion...
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  7. #7
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    In the integral is the composition x \circ u^{3} and the composition is the multiplication.

    What has the solution of u to do with the function x. Thats my problem. That are two functions and why do they correlate?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Besserwisser View Post
    In the integral is the composition x \circ u^{3} and the composition is the multiplication.

    What has the solution of u to do with the function x. Thats my problem. That are two functions and why do they correlate?
    Thinking about composite functions is unnecessary for completing the task at hand.

    When we integrate, our objective is to make the integrand easy to work with. Here's what I'm gonna do:

    I'm gonna do this integral from scratch, and then ou tell me when you get lost.

    1. Problem: \int{x}\sqrt{x+1}dx

    2. Notice that making the substitution u=\sqrt{x+1} might get me somewhere, because as it stands now, this particular integral is nasty. The key is to TRY SOMETHING. Anything at all.

    3. If u=\sqrt{x^2+1} then by solving for x I obtain x=u^2-1.

    4. Differentiate \frac{dx}{du}=2u which implies that dx=2udu

    5. Look at the original problem and make the substitutions

    \int\overbrace{x}^{u^2-1}\overbrace{\sqrt{x+1}}^u\overbrace{dx}^{2udu} giving us

    \int(u^2-1)u2udu

    So, where is your trouble. Step 1? 2? 3? etc...
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  9. #9
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    I have only with one word a problem. In step three you say "obtain". If I obtain I got everything. "Obtain" is the problem.

    The rule of substitution in general says:

    Is \varphi : [a,b] \rightarrow \mathbb{R} continous differentiable and I \subset \mathbb{R} an intervall with  \varphi([a,b]) \subset I and f: I \rightarrow \mathbb{R} piecewise continous, than is

      \int\limits_{\varphi(a)} ^{\varphi(b)} f(x) \mathrm{d}x = \int\limits_{a}^{b} f(\varphi(t)) \cdot \varphi \prime (t) \mathrm{d}t

    I understand the scentence and the proof of it but I can not apply it to the example above. The other x is not an argument of u, so why we can obtain this? We know nothing about the area of definition here.

    Argh, I guess I am too stupid to see the answer
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Besserwisser View Post
    I have only with one word a problem. In step three you say "obtain". If I obtain I got everything. "Obtain" is the problem.

    The rule of substitution in general says:

    Is \varphi : [a,b] \rightarrow \mathbb{R} continous differentiable and I \subset \mathbb{R} an intervall with  \varphi([a,b]) \subset I and f: I \rightarrow \mathbb{R} piecewise continous, than is

     \int\limits_{\varphi(a)} ^{\varphi(b)} f(x) \mathrm{d}x = \int\limits_{a}^{b} f(\varphi(t)) \cdot \varphi \prime (t) \mathrm{d}t

    I understand the scentence and the proof of it but I can not apply it to the example above. The other x is not an argument of u, so why we can obtain this? We know nothing about the area of definition here.

    Argh, I guess I am too stupid to see the answer
    Dude, you're thinking way too hard.

    From intermediate algebra (not advanced calculus) we "obtain"...

    u=\sqrt{x+1}.

    Squaring both sides

    u^2=(\sqrt{x+1})^2

    so...

    u^2=x+1

    Subtracting 1 from both sides we "obtain"

    x=u^2-1

    fin
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  11. #11
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    But I have to study "advanced calculus" (not my wish, it is part of my required courses) and want to apply the scentence. I want to understand the reasons why we do this calculation and I didn't see it
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Besserwisser View Post
    But I have to study "advanced calculus" (not my wish, it is part of my required courses) and want to apply the scentence. I want to understand the reasons why we do this calculation and I didn't see it
    "Oh No! There has been a breech of security! Ever body get down." A voice screamed.

    In, the backround, the machine guns raged on. And poor Budweiser went down down. He went down hard.

    "Poor guy," I thought. "He didn't even get to see the big picture."



    Story dedicated to the general, you know who you are.
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  13. #13
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    Quote Originally Posted by Besserwisser View Post
    But I have to study "advanced calculus" (not my wish, it is part of my required courses) and want to apply the scentence. I want to understand the reasons why we do this calculation and I didn't see it
    OK.
    You have an integral in terms of "x".
    and you did a U-Substitution.
    then you should convert everything in the original integral in the corresponding value in terms of u according to your substitution.
    VonNemo19 explained it two times, then he explained it as a steps.
    and you have just moved on a circular.
    I tried to explain it to you 5 times, But really I can not.
    Do you know why?
    Because it is clear !
    do not ask yourself this kind of questions.
    try to ask yourself -as an example- " What if it is \sqrt{x^3+1}, Can I integrate it ?
    What If it was tan(x) there, Can I integrate it ?
    These kind of questions will help you.
    such questions will let you bring your pen and your papers and do a "GOOD TRY", they will learn you new methods for integrating things.
    Did you see the difference between them ?
    Your question makes you moving around a circular.
    but the other questions let you learn new methods .
    Come on, There is many people on our earth who can not just pass Calculus I !
    Sure you are better than them.
    always try to search about the "white" things, not the "black" things.
    Sorry, but this thread has "no goal" !
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