Results 1 to 7 of 7

Math Help - Need help with this integral

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    55

    Need help with this integral

    Hi guys

    I've been doing some practice problems on integration and this particular one stumped me.

    Int(|sqrt(x)-1|.dx)

    Evaluate the integral at (0,4).

    So far I've gotten this when x>1, sqrt(x)-1>0, so the integral is 2/3.x^(3/2)-x+C. I'm still stuck on the (0,1) portion.

    I'm a little rusty at integration so any advice would be appreciated.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Notice that on the interval [0,1], sqrt(x) - 1 is less than 0 or equal to zero, and from [1, 4] it is greater than or equal to zero. Then you can split the integral into two, and then get rid of absolute value. Remember that:

    \int^{b}_{a} f(x)~dx =\int^{c}_{a} f(x)~dx + \int^{b}_{c} f(x)~dx

    Provided that c \in (a, b)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    55
    Yea I've tried splitting them up. Would it be right to say that on (0,1) it can be integrated as -sqrt(x)+1?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by Hweengee View Post
    Yea I've tried splitting them up. Would it be right to say that on (0,1) it can be integrated as -sqrt(x)+1?
    Yes.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    55
    ah ok thanks. that makes it much clearer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2008
    Posts
    55
    Ok this may sound weird, but what happens when x<0? the root of a negative number is complex so what happens to the absolute value of a complex number minus 1?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Hweengee View Post
    Ok this may sound weird, but what happens when x<0? the root of a negative number is complex so what happens to the absolute value of a complex number minus 1?
    In that case, consider the absolute values signs as modulus :

    |a+ib|=\sqrt{a^2+b^2}

    Otherwise, you can use the complex logarithms, but it's not what you have here
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 08:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum