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You Can't Get There
Improper Integrals

 

 When we have talked about definite integrals,

 the limits of integration 
 (the numbers on the top and bottom of the funny S)
 have always been real numbers, like p or 12 or whatever.
 But what if we wanted to use something like infinity ()  for the upper limit
 
 You might think that any time we did that, 
 the answer that you would get would be infinity.
 
 Sometimes you do, but not always.
 
 For example, if you walked half way across a room
 and then walked half of the remaining way
 and then walked half of the distance that was left.
 If you keep on walking half the remaining distance each time
 you can take as many turns as you want
 but you will never get to the other side of the room.
 
 This is like integrating your step turns from one to infinity.
 The total distance you walk will not be infinity.
 It would keep getting closer to the distance across the room
 but never quite get there.
 
 The deal is, if the distance you move on each step
 gets smaller each step "at a fast enough rate"
 the distance traveled will approach some number.
 
 If the distance you move on each step
 does not get smaller "at a fast enough rate"
 the distance traveled will approach infinity.
 
 Hey, I bet you could write that as some kind of limit problem!
 
 Well, now that you mention it ...
 
 Say we have some function like ...
 

 
 The graph of this looks like ...
 

 
 and suppose that we want to find the area 
 between the graph line and the X axis 
 from the point where X = 1 to X = .
 

 
 To begin with, the problem would look like ...
 

 
 Now we could just integrate and see what we get ...
 

 
 Now that negative one over two times infinity term is pretty weird.
 Working with it in official math takes a bit of work,
 but it's obvious to most people that ...
 

 
 So our area is equal to that -(-1/2) term.
 The term simplifies to 1/2,
 but the official, get all the points on the exam solution,
 looks like this ...
 

 

 
 You probably noticed that the graph has a vertical asymptote
 at the point where X = 0.
 
 What of we wanted to see if we could calculate the area
 from X = 0 to X = 1?
 

 
 We got a number for the area under the graph
 from X = 1 to X = even though the graph area goes forever.
 Maybe this unbounded area from X = 0 to X = 1
 will give us a finite (some real number value) area.
 Let's try ...
 

 
 Nope, the area calculates as infinity.
 
 Interesting, the 2 areas look about the same
 but one has an infinite are and the other has a finite area.
 
 Let's try another one ...
 

 
 Let's calculate the area of this one from X = 0 to X = 1.
 That's the part that blew up in the last example.
 

 
 Wow! It worked that time.
 Now let's try the X = 1 to X = part.
 

 
 Hmmm,
 On this one, where the exponent was - 1/2 
 the part where X went to infinity blew up.
 On the one where the exponent was -2
 the part where X went from zero blew up.
 
 I wonder if we could generalize and say 
 that for exponents between -1 and 0
 we get an infinite area when we integrate
 from X equals some number to X equals infinity.
 And a finite area when we integrate 
 from X equals zero to X equals some finite number.
 
 And if it also always holds that when the exponent is less than -1
 the reverse is true.
 

 
 YES! 
 THIS IS EXACTLY THE WAY IT WORKS!
 
 I wonder what happens at the case between these two.
 That is, at ...
 

 F(X) = X-1

 

 
 Let's try it (both parts)
 

 
 Oh well, the area of both parts is infinite.
 
 Two comments.
 First, the lim on the right used "b" instead of "a" to avoid confusion.
 Second, if you can't see how the limit of ln a as a approaches 0 is infinity,
 calculate ln 0.5, then ln 0.005, then ln 0.00005.
 

   copyright 2005 Bruce Kirkpatrick

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