# Improper integrals

1. Apr 4, 2009

### lha08

1. The problem statement, all variables and given/known data
Determine whether the integral is divergent or convergent. If it is convergent, evaluate it.
Integral from (1 to 8) 4/(x-3)^3 dx

2. Relevant equations

3. The attempt at a solution
So firstly, i pulled out the 4 and then i set the lim as t goes to 8 from the left of the integral. From there, i applied the u sub with u=x-3 and du=dx. I then did the antiderivative of it which gives me: 4 lim (as t goes to 8 from left) integral from 1 to t (1/(-2u^2)) and then i changed the limits of integration...t=t-3 and 1=-2. but when i plugged them in and evaluated i got 0.42..which is wrong...i don't know where i'm going wrong. thanks.

2. Apr 4, 2009

### Dick

Are you sure it's convergent? What happens at x=3?

3. Apr 4, 2009

### lha08

ohh the bottom heads the zero so at x=3 it does not exist, right? so it diverges..

4. Apr 4, 2009

### Dick

In this case, yes, it diverges. But just having a zero denominator doesn't prove that. You are dealing with improper integrals. 1/sqrt(x) has a finite integral on [0,1]. Why is your case different?

5. Apr 4, 2009

### lha08

Um i'm not really sure but the first thought that comes to mind is that 4/0 does not exist and that 3 is found in the limits of integration..but i'm really not sure...

6. Apr 4, 2009

### Dick

You are doing 'improper integrals', right? To figure out whether 1/(x-3)^3 has an integral on [0,3], don't you do the integral from 0 to 3-e where e is some small positive number and then take the limit as e tends to zero?

7. Apr 4, 2009

### n!kofeyn

You are right that 3 is in the interval [1,8] and that the function has a discontinuity at x=3. Do you know of a technique that can get the point 3 in the limits of integration?

Note that in the case of $1/\sqrt{x}$, integrated over the interval [0,1], the discontinuity happens at an endpoint of the interval. Then the integral can be evaluated by:
$$\int_0^1 \frac{1}{\sqrt{x}} \,dx = \lim_{t\to 0^+} \int_t^1 \frac{1}{\sqrt{x}} \,dx$$