# Homework Help: 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.