For problem 5 part b, why is it that during the third step of the integral you change e^-2(x-L/2)/a to be e^-(2x/a), and the limits of integration are now 0 -> ∞?
That is a change of variables (by a translation) from x to a new x=(x-L/2). You would get the same thing with or without that change.
really, i should have introduced a new variable u=x-L/2, and du = dx. (and then integrated from u= 0 to infnty, which corresponds to x = L/2 to infnty. Does that make more sense?
just to simplify. it wasn't really necessary at all. i kind of got in the habit of putting integrals into W-A, even when i didn't need to (like this one).
For problem 5 part b, why is it that during the third step of the integral you change e^-2(x-L/2)/a to be e^-(2x/a), and the limits of integration are now 0 -> ∞?
ReplyDeleteThat is a change of variables (by a translation) from x to a new x=(x-L/2). You would get the same thing with or without that change.
ReplyDeletereally, i should have introduced a new variable u=x-L/2, and du = dx. (and then integrated from u= 0 to infnty, which corresponds to x = L/2 to infnty. Does that make more sense?
Yes thinking of it with a COV using u makes sense. Did you do this just to simplify the integral, or was there some other motivation?
ReplyDeletejust to simplify. it wasn't really necessary at all. i kind of got in the habit of putting integrals into W-A, even when i didn't need to (like this one).
Delete