# Improper integrals

1. Feb 24, 2014

### Chas3down

1. The problem statement, all variables and given/known data

integral of 1/sqrt(9-x^2)
from 0 to 3

2. Relevant equations

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3. The attempt at a solution
I integrate it correct to arcsin(x/3) from 0 to 3
Get the correct anwser of pi/2.

But there is another question, At which value of x in the integration region [0,3] does special care need to be taken with the integration? I understand at some point it goes from negatie to positive, but i tried 0,3,pi/2,pi.. none worked.. anyhelp?

2. Feb 24, 2014

### LCKurtz

What did you get when you put $x=3$ into the integrand?

And why do you say it goes from negative to positive?

3. Feb 24, 2014

### scurty

Check the domain of the original function to be integrated.

4. Feb 24, 2014

### Chas3down

I never had to do anything with the domain, it just worked.. But i guessed 0, pi/2 and 3. I thought it was be pi/2 because thats where it goes from neg to pos.

5. Feb 24, 2014

### scurty

That's because you have been doing "proper" integrals up to this point. Improper integrals involve integrating across a point where the function is not defined. In this case the the function is not defined at x = __. The normal procedure is to introduce a variable for that number and take the limit as a approaches that number.

In this case, arcsin is defined on [0,1]. But the original function is not defined on [0,3].

6. Feb 24, 2014