Results 1 to 6 of 6

Math Help - Integration via substitution method

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    1

    Integration via substitution method

    Hi,
    Im learning to do integration using the substitution method.
    My problem is:
    Integrate
    x/[root(1+2x)]dx
    (4,0) (the 4 and 0 are at the top and bottom of the integration sign (the upper and lower bound)
    using the substitution method.
    i let u = 1+2x
    so du = 2dx (?)
    then i tried substituting back into the equation:
    dx = 1/2du
    so..


    Any help would be appreciated.. i dont know if this is right.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2008
    Posts
    128
    \int_0^4 \sqrt{1+2x} dx

    let  u = 1+2x then  du=2dx so  \frac{1}{2}du=dx

    You can change your x-limits of integration to u-limits if you would like, or you could keep them in terms of x. I prefer to keep them in terms of x but if you were to change them to u-limits, you would plug 0 =x and 4=x into the u equals equation.

    \int_{x=0}^{x=4}\frac{1}{2}\sqrt{u} du

    Can you finish from here? Integrate, put u back in terms of x, then evaluate from zero to four.

    PS: This is exactly what you did. U-sub can be intimidating at first, but after a couple of weeks it will seem a lot easier. Trust yourself enough to make mistakes. You were on the right track EXACTLY! I bet if you had kept going you would have definitely gotten it..because I didn't do anything different from what you did in your first posting.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    A more direct sub. is to set u^2=1+2x.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2008
    Posts
    128
    Quote Originally Posted by Krizalid View Post
    A more direct sub. is to set u^2=1+2x.
    Yes, but I find when I'm first trying to explain u-sub to the students I tutor, their minds can't make that leap. I think you first have to master the basics before you can jump into "seeing" the different things you could let u equal. So by letting plain old u equal what is under the radical it's a simple variable that you can manipulate and still integrate easily...I think that letting everything equal u^2 would actually be harder than just simply integration square root of u for a beginner. Any thoughts?

    Hopefully that was coherent, it's been a long day.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    Then write it backwards as u=\sqrt{1+2x}, so I think there's no problem with that.

    Before seein' these stuff the student must have a full domain of differentiation techniques.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2008
    Posts
    128
    Quote Originally Posted by Krizalid View Post
    Then write it backwards as u=\sqrt{1+2x}, so I think there's no problem with that.

    Before seein' these stuff the student must have a full domain of differentiation techniques.

    Touche! I see your point. Thank you for the enlightening posting..perhaps I will point this out to my little calc I cherubs when we talk about u-sub.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral Method using the Substitution Method
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 16th 2011, 07:10 PM
  2. using the substitution method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2009, 09:43 AM
  3. Replies: 2
    Last Post: May 2nd 2009, 09:14 PM
  4. Method of Substitution
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 1st 2009, 07:47 AM
  5. Method of Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 1st 2006, 04:56 PM

Search Tags


/mathhelpforum @mathhelpforum