Results 1 to 4 of 4

Math Help - integration problems

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    32

    integration problems

    could someone help me integrate the following equations because i keep getting the wrong answer and i dont no why T.T thankyou !

    a) x/[(2x-1)^1/2]
    b) dt/dx = 1/{2[(4-x)^2]} and that x = 0 when t=0 find x in terms of t

    btw i am using the u substitution method for these two questions

    thankyou so much for any help given !! =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Let u=2x-1 so du=2dx

    So  \int {xdx\over \sqrt{2x-1}} ={1\over 4} \int {(2x)(2dx)\over \sqrt{2x-1}}

     ={1\over 4} \int {(u+1)du\over \sqrt{u}} ={1\over 4} \int (u^{1/2}+u^{-1/2})du
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by flyinhigh123 View Post
    could someone help me integrate the following equations because i keep getting the wrong answer and i dont no why T.T thankyou !

    a) x/[(2x-1)^1/2]
    b) dt/dx = 1/{2[(4-x)^2]} and that x = 0 when t=0 find x in terms of t

    btw i am using the u substitution method for these two questions

    thankyou so much for any help given !! =)
    For a), you can use integration by parts.

    Remember that \int{u\,dv} = uv - \int{v\,du}.


    \int{\frac{x}{(2x - 1)^{\frac{1}{2}}}\,dx} = \int{x(2x - 1)^{-\frac{1}{2}}\,dx}.


    Let u = x so that du = 1.

    Let dv = (2x - 1)^{-\frac{1}{2}} so that v = (2x - 1)^{\frac{1}{2}}.


    So \int{\frac{x}{(2x - 1)^{\frac{1}{2}}}\,dx} = \int{x(2x - 1)^{-\frac{1}{2}}\,dx}

     = x(2x - 1)^{\frac{1}{2}} - \int{(2x - 1)^{\frac{1}{2}}\,dx}

     = x(2x - 1)^{\frac{1}{2}} - \frac{1}{3}(2x - 1)^{\frac{3}{2}} + C.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by flyinhigh123 View Post
    could someone help me integrate the following equations because i keep getting the wrong answer and i dont no why T.T thankyou !

    a) x/[(2x-1)^1/2]
    b) dt/dx = 1/{2[(4-x)^2]} and that x = 0 when t=0 find x in terms of t

    btw i am using the u substitution method for these two questions

    thankyou so much for any help given !! =)
    \frac{dt}{dx} = \frac{1}{2(4 - x)^2}

     = \frac{1}{2}(4 - x)^{-2}

     = -\frac{1}{2}(4 - x)^{-2}(-1).


    Let u = 4 - x so that \frac{du}{dx} = -1


    So

    t = \int{ \frac{1}{2(4 - x)^2}\,dx}

     = -\frac{1}{2}\int{(4 - x)^{-2}(-1)\,dx}

     = -\frac{1}{2}\int{u^{-2}\,\frac{du}{dx}\,dx}

     = -\frac{1}{2}\int{u^{-2}\,du}

     = \frac{1}{2}u^{-1} + C

     = \frac{1}{2(4 - x)} + C.


    Now use your initial condition to find C.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2010, 06:39 PM
  3. integration problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2009, 05:01 AM
  4. Integration problems
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 26th 2008, 01:02 AM
  5. Problems with some integration
    Posted in the Calculus Forum
    Replies: 21
    Last Post: July 25th 2006, 04:16 AM

Search Tags


/mathhelpforum @mathhelpforum