00:01
We use the wave function for a particle in a box for any value in of the quantum number.
00:07
So we'll go ahead and calculate as follows.
00:15
So we'll end up squaring the wave function, so we have 2 over l outside of the integral.
00:22
And we're integrating on the interval from 0 to l.
00:27
So i should once again, i should state that this interval is really only for 0 to l.
00:34
And the wave function is zero elsewhere.
00:41
So this wave function equals zero anywhere outside of this interval.
00:47
So this is zero elsewhere.
00:53
And continuing with what we have here, we'll write x squared, sine squared, n pi x over l, dx.
01:15
And we'll go ahead and write on the next line here.
01:21
This is going to be equal to 2 over l, and in brackets we'll have x cubed over 6 minus, in parentheses, i'm going to write x squared over 4 times l over n pi minus l cubed over 8 pi cubed, n cubed.
02:03
This is all in the denominator right here.
02:06
And then we'll close off these parentheses.
02:12
And then this multiplies sine 2 pi nx over l.
02:24
And then we have minus l squared over times x over 4 pi squared, n squared, cosine of 2 pi nx over l and finally we close our square brackets and we evaluate on the interval zero to l.
02:59
So there's a lot of things going on here, as you can see, but this is going to greatly simplify and i'm going to explain why, because this sign is going to go away.
03:12
This is going to help us out a lot.
03:14
When i look at l and i look at 0 and i substitute 0 or l in for x right here, i'm going to end up with, let's say that i put 0 in.
03:27
I'm going to get sign of 0.
03:29
So on the lower point of the interval, the lower end point, i get sign 0.
03:35
If i put in l, i'm going to have 2 pi in l, l over l, which gives me sign of 2 pi in...