Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Verification neededproblem seems odd

  1. Oct 7, 2007 #1
    Hi,

    I have to find the integral of 1/square root(x-3)dx

    this is what i did

    let u=sqr root(x-3)
    du=1/2*square root(x-3)dx
    dx=2*square root(x-3)du

    the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

    ..ok now what i did is plug back u= square root of(x-3) in the denom
    that way Im only left with 2du.

    Then i integrated that to 2x2/2 which is equal to x^2.

    Ultimatley I got the integral to equal X !!!! which cant be right!!!

    But doesnt this make sense, I know i did not do anything wrong...
     
  2. jcsd
  3. Oct 7, 2007 #2
    [tex]\frac{d}{dx}\left(\sqrt{x-3}\right) \neq \frac{1}{2}\left(\sqrt{x-3}\right)[/tex]
     
  4. Oct 7, 2007 #3
    ..ok now what i did is plug back u= square root of(x-3) in the denom
    that way Im only left with 2du.

    that move is NOT legit.

    the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

    neither is that one, wen u change variables, ur new equation must have only that variable - or constants of course...if you write root(x-3) IN TERMS of u that is fine, but the way it is written doesn't make sense.

    Hint: let u=x-3.
     
  5. Oct 7, 2007 #4
    I meant that the derivitive of sqroot(x-3)=1/2sqroot(x-3)
     
  6. Oct 7, 2007 #5
    And I'm saying that it is not. :wink:

    [tex]\frac{d}{dx}\left({x-3}\right)^\frac{1}{2} = \frac{1}{2}\left(\frac{1}{\sqrt{x-3}}\right)[/tex]
     
  7. Oct 7, 2007 #6
    "1/2sqroot(x-3)", i think this is 1 OVER 2 times the square root of x minus 3.
     
  8. Oct 7, 2007 #7
    Your first part is flawless. Integrate by parts:
    [tex]\int xdu = xu-\int udx[/tex]
    set [tex]x=2[/tex] and [tex]u=\sqrt{x-3}[/tex]. Therefore [tex]du=\frac{1}{2\sqrt{x-3}}[/tex]. Obviously [tex]dx=0[/tex].
    Plug in your values:
    [tex]\int 2(\frac{1}{2\sqrt{x-3}}) = 2(\sqrt{x-3})-\int 0[/tex]

    Simplify and you get:
    [tex]\int \frac{1}{\sqrt{x-3}} = 2\sqrt{x-3}[/tex]

    As you do more integrations, this process will become easier.
     
  9. Oct 7, 2007 #8
    u-substituting x-3 is a lot easier!
     
  10. Oct 7, 2007 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The problem is that most people would interpret it as (1/2) sqroot(x-3), not 1/(2 sqroot(x-3)). USE PARENTHESES!
     
  11. Oct 7, 2007 #10
    that he didn't make that mistake is clear. Look at his next step:

    dx=2*square root(x-3)du
     
  12. Oct 8, 2007 #11
    Ah... I apologise, gochi. But as Halls emphasised, please make use of parentheses. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Verification neededproblem seems odd
  1. Verification of Method (Replies: 2)

  2. Seemingly easy integral (Replies: 15)

Loading...