Improper Integration

1. Apr 17, 2010

temaire

The problem statement, all variables and given/known data

3The attempt at a solution

Is this correct?

2. Apr 17, 2010

Dick

Sure, it looks fine to me.

3. Apr 17, 2010

temaire

Do I need to show that I'm approaching 4 from the right?

4. Apr 17, 2010

Dick

I would say you are approaching 4 from the left. t<4, correct? Why would you want to approach from the right?

5. Apr 17, 2010

temaire

Because if I were approaching ln|x-4| from the right, the graph goes to -∞. How could you approach it from the left?

Or am I supposed to approach the original graph of 1/(x^2 -3x -4) from the left?

6. Apr 17, 2010

Dick

The problem with your integral is at x=4. To resolve it as an improper integral you want to integrate from x=0 to x=4-epsilon where epsilon>0. That means you are approaching the upper limit from the left. You don't care what the limit is from the right.

7. Apr 17, 2010

temaire

So even though the limit is approaching 4 from the left at ln|x-4|, we're infact evaluating the limit as it approaches 4 from the left of 1/(x^2-3x-4)?

8. Apr 17, 2010

Dick

Now you are just confusing me. You are approaching x=4 from the left. Period. Approaching ln|x-4| from the left gives you the behavior of the integral of 1/(x^2-3x-4) on the interval [0,4].

9. Apr 18, 2010

temaire

This is what I was confused about.

In the graph above, you can approach 4 from the right but not from the left.

10. Apr 18, 2010

Dick

It's an absolute value, temaire. |x-4|. Doesn't that mean anything to you? :)

11. Apr 18, 2010

temaire

Oh, so this is the graph. (x is from 3 to 5)

I understand now.

12. Apr 18, 2010

Dick

You've got it.