Results 1 to 9 of 9

Math Help - Evaluate the integral

  1. #1
    Senior Member
    Joined
    Jul 2007
    Posts
    290

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    There's a discontinuity at x=2, but I think that those aren't the right bounds.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by JonathanEyoon View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    *sigh* Appreciate it!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Krizalid View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    'cause you get a negative square root, it makes no sense.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Does that mean the problem itself can't be solved? That would make my day . If so, can you tell me how?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by JonathanEyoon View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Yes sir! Thanks I appreciate it guys. As always you guys are extremely helpful!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 1
    Last Post: November 28th 2009, 08:44 AM
  4. Evaluate the integral using the FTC I
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 7th 2009, 05:29 AM
  5. How do you evaluate this integral?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 15th 2008, 04:12 AM

Search Tags


/mathhelpforum @mathhelpforum