Thread: Schrodinger:Normalise Psi- simple integration query

1. 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

2. 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.

3. 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.

4. here guys,

see attached. Must have made a mistake somewhere

thanks

5. $\displaystyle\int b^{2}\,dx=b^{2}x,$ not $b^{3}/3.$ That's part of the problem.

6. Originally Posted by Ackbeet
$\displaystyle\int b^{2}\,dx=b^{2}x,$ 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 :-)

7. Originally Posted by bugatti79
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.

8. nope! still stuck. i think i should give up maths and read childrens books

9. Re-define the constant of integration. Then you can get your b^3.

10. Originally Posted by bugatti79
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.

11. 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!! :-)

12. Originally Posted by Ackbeet
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

13. Originally Posted by bugatti79
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).

14. Originally Posted by mr fantastic
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