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Math Help - Schrodinger:Normalise Psi- simple integration query

  1. #1
    Senior Member bugatti79's Avatar
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    Schrodinger:Normalise Psi- simple integration query

    Hi All,

    I have a query regarding a simple integration problem. See attached. It is regarding integrating (b-x)^2dx. The correct way is integrating this as a whole. At a first glance, I would have multiplied out the brackets and then integrated.....

    Thanks
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  2. #2
    Master Of Puppets
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    Use this approach

    \displaystyle \int (ax+b)^n~dx = \frac{(ax+b)^{n+1}}{a(n+1)}+C

    Which way you integrate gives you the same answer, post the two answers, i'll show you how.
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  3. #3
    A Plied Mathematician
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    Or just do a u substitution: u = b-x, du = - dx. The point is, doing it this way saves you the trouble of multiplying it all out, integrating term-by-term, and then re-factoring to simplify again.
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  4. #4
    Senior Member bugatti79's Avatar
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    here guys,

    see attached. Must have made a mistake somewhere

    thanks
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  5. #5
    A Plied Mathematician
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    \displaystyle\int b^{2}\,dx=b^{2}x, not b^{3}/3. That's part of the problem.
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    \displaystyle\int b^{2}\,dx=b^{2}x, not b^{3}/3. That's part of the problem.
    I still dont see where else I am making mistake... I find it frustrating that I seem to be making simple msitakes! Is that common among people or is it just me :-)
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  7. #7
    Flow Master
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    Quote Originally Posted by bugatti79 View Post
    I still dont see where else I am making mistake... I find it frustrating that I seem to be making simple msitakes! Is that common among people or is it just me :-)
    Go back and do the problem from scratch, keeping mind post #5.
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  8. #8
    Senior Member bugatti79's Avatar
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    nope! still stuck. i think i should give up maths and read childrens books
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  9. #9
    A Plied Mathematician
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    Re-define the constant of integration. Then you can get your b^3.
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  10. #10
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    Quote Originally Posted by bugatti79 View Post
    nope! still stuck. i think i should give up maths and read childrens books
    You have calculated a correct anti-derivative. Use it to get the required definite integral. Add the results of the two integrals together. Equate to 1. Solve for A. Done.
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  11. #11
    Senior Member bugatti79's Avatar
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    Fantastic, I understand the original 'normalise psi question'. Its query I have regarding 2 ways of integrating (b-x)^2
    1) using pickslides or ackbeets method which is fine or
    2) my method of expanding out the brackets and integrating.

    ok guys, I will look at it tomorrow. Its saturday night, I will mull it over my pint of guinneas!! :-)
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  12. #12
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Re-define the constant of integration. Then you can get your b^3.

    Not sure what you mean by this. We dont need any definite integrals here (I think).
    integral of (b-x)^2 = b^2x-bx^2+x^3/3+C....
    Thanks
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  13. #13
    Flow Master
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    Quote Originally Posted by bugatti79 View Post
    Not sure what you mean by this. We dont need any definite integrals here (I think).
    integral of (b-x)^2 = b^2x-bx^2+x^3/3+C....
    Thanks
    My previous post tells you what to do. Do it. (There are several ways to integrate - they all lead to equivavlent antiderivatives, but that is irrelevant because you are required to do a definite integral).
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  14. #14
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by mr fantastic View Post
    My previous post tells you what to do. Do it. (There are several ways to integrate - they all lead to equivavlent antiderivatives, but that is irrelevant because you are required to do a definite integral).
    Thanks Mr Fantastic.

    I got myself in a twist because I came across the same problem in a different application. It turns out that if I correctly follow rules regarding the constants and limits the 3 methods work the same as you said. Thanks also to ackbeet and pickslides. See attached summary.

    Cheers
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