Solving Improper Integrals: 1/sqrt(9-x^2) 0 to 3

[Chas3down]

Homework Statement

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

Homework Equations

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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?

 

[LCKurtz]

Homework Statement

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

Homework Equations

///

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?

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

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

 

[scurty]

Homework Statement

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

Homework Equations

///

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?

Check the domain of the original function to be integrated.

 

[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 that's where it goes from neg to pos.

 

[scurty]

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 that's where it goes from neg to pos.

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].

 

[LCKurtz]

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

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

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 that's where it goes from neg to pos.

Try answering my two questions.