1. ## Evaluate the integral

x / (sqrt (x^2 - 4)) from 0 to 4

I chose to use trig substitution to do this problem but realized i couldn't set a lower boundary in terms of theta for 0......or...can I?

THEN, I realized i could use u substitution..... anyways when I used that, I quickly came to a part where I would have to find the square root of -4 which... doesn't give me a real number answer. Can someone shed some light onto me for this problem?

I'm gonna go ahead and throw it out there, but would this by chance be an improper integral? Thanks

2. There's a discontinuity at $x=2,$ but I think that those aren't the right bounds.

3. Originally Posted by JonathanEyoon
x / (sqrt (x^2 - 4)) from 0 to 4

I chose to use trig substitution to do this problem but realized i couldn't set a lower boundary in terms of theta for 0......or...can I?

THEN, I realized i could use u substitution..... anyways when I used that, I quickly came to a part where I would have to find the square root of -4 which... doesn't give me a real number answer. Can someone shed some light onto me for this problem?

I'm gonna go ahead and throw it out there, but would this by chance be an improper integral? Thanks
Yes, a divergent improper integral

4. *sigh* Appreciate it!

5. Originally Posted by Krizalid
There's a discontinuity at $x=2,$ but I think that those aren't the right bounds.

What do you mean by those aren't the right bounds?

6. 'cause you get a negative square root, it makes no sense.

7. Does that mean the problem itself can't be solved? That would make my day . If so, can you tell me how?

8. Originally Posted by JonathanEyoon
x / (sqrt (x^2 - 4)) from 0 to 4

I chose to use trig substitution to do this problem but realized i couldn't set a lower boundary in terms of theta for 0......or...can I?

THEN, I realized i could use u substitution..... anyways when I used that, I quickly came to a part where I would have to find the square root of -4 which... doesn't give me a real number answer. Can someone shed some light onto me for this problem?

I'm gonna go ahead and throw it out there, but would this by chance be an improper integral? Thanks
$\sqrt{x^2-4}=f(x)$

f(x)'s domain is $(-\infty,-2)\cup(2,\infty)$

So any function of the form $f(g(x))$, or in other words any function containing f(x) will not be defined $\forall{x}\in[-2,2]$

Make sense now?

9. Yes sir! Thanks I appreciate it guys. As always you guys are extremely helpful!