2007-09-26

Improper integrals

• 1A. When the limit of integration is infinite, consider if you would: the integral from 1 to infinity of (1/x^2)dx. Well, we can't do this. We can't go to infinity. However, we need to approach this as a limit. So, infinity can't be there. So we're going to put something in front:
  limit [as B->infinity] of integral [from 1 to B] of (1/x^2)dx. ===> limit [as B->infinity] of -x^{-1}. FROM 1 to B.
  = Plug in the upper limit. That's lim [B->infinity] of (-(1/B) - (-1)). What does this equal as B->infinity? So the limit is 1.
  We say that the improper integral (integral [from 1 -> infinity] of 1/x^2 dx) converges to 1.
  
  -- You must rewrite the initial problem as a limit of an integral and consider as x approaches infinity. So on the AP test you have to include that. You have to have a setup to get your credit. Improper integrals, sometimes can be evaluated in a sense, and sometimes they can't be. What's improper about them? That's what we need to look at.
  
  
  
• 1B. Now, consider the integral [from 1 -> infinity] of (1/x)dx. So that's:
  lim [as B->infinity] of integral [from 1 to B] of (1/x)dx.
  lim [as B->infinity] of ln(x) FROM 1 to B.
  lim [as B->infinity] of (lnB - ln1).
  -- So you have infinity-minus-zero and the limit of this, well, in this case, we say the improper integral diverges.