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Integral of a Sqrt

  1. Oct 14, 2003 #1
    I try to take the integral of sqrt(4x-1) with respect to x...


    The correct answer is sqrt((4x-1)^3)/6, but I always get

    2sqrt((4x-1)^3)/3... Can someone explain how to solve that integral pls.
     
  2. jcsd
  3. Oct 14, 2003 #2

    Tom Mattson

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    OK

    No, it isn't.

    Edit: My mistake; yes it is.

    That's not right either.

    Yes, you do a u-substitution. Let u=4x-1, so du=4dx. You then have:

    (1/4)∫u1/2du,

    which is elementary.
     
    Last edited: Oct 14, 2003
  4. Oct 17, 2003 #3
    If you didn't understand what the last person said when you take the integral of squrt(4x-1) with respect to xdx you must first find u and u prime (or the derivative of u), where u is what is in the parentheses, in this case (4x-1) so u prime (or the derivative of (4x-1)) = 4x, but you still have to deal with the remaining x and the 4 left by the derivative. To get rid of the 4 in the derivative you divide the integral by 4, and subistute for the x so. x=(u-1)/4, I'm you understand the rest. You multiply u^(1/2)*((u-1)/4) then take the antiderivative of that.
     
  5. Oct 17, 2003 #4

    HallsofIvy

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    The problem, by the way, is NOT the square root! You can easily integrate ∫ √(x) dx. √(x)= x1/2 and you can use the "power rule".
    The problem is that "4x-1" inside the square root. To get rid of that you make the substitution mentioned earlier: u= 4x-1 so that
    √(4x-1)= &radi;(u)= u1/2. Of course, you have to convert from "dx" to "du". Because 4x-1 is linear, that's easy
    du/dx= 4 so du= 4 dx or (1/4)du= dx.

    ∫ &radic(4x-1)dx= (1/4)&int u1/2du.

    The power rule says that an anti-derivative of un is
    1/(n+1) un+1. In this case, n= 1/2 and n+1= 3/2. The anti-derivative is (2/3)u3/2+ C . Replacing u by 4x-1 again, ∫ &radic(4x-1)dx= (2/3)(4x-1)3/2+ C.

    Since the original problem was given in terms of √ rather than a 1/2 power, it might be a good idea to set the answer in those terms: ∫ &radic(4x-1)dx= (2/3)(√(4x-1))3+ C.
     
  6. Oct 19, 2003 #5
    Why can't you just int. the bracket like you would int. x^0.5?

    Code (Text):

    ∫ (4x-1)^0.5 dx

    increase the power and divied by it:

    (4x-1)^1.5 + c = 2(√(4x-1))^3 + c
    ----------       ------------
       1.5                3
     
     
    Last edited: Oct 19, 2003
  7. Oct 19, 2003 #6
    Because if you differentiate your result ((2/3)(4x-1)^(3/2) + C) you get:

    4(4x - 1)(1/2)

    which is not what you integrated.
     
  8. Oct 19, 2003 #7

    Tom Mattson

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    As Sting said, when you differentiate your result, you do not get the original integrand. That is because (4x-1)1/2 is a composition of functions.

    f(u)=u1/2
    u(x)=4x-1

    In general, a composition of functions does not satisfy the same basic integration rule as the simple function f(u).

    edit: typo
     
  9. Oct 19, 2003 #8
    Sorry, I meant to integrate it as a sight integral, where a multiple of the differential of the bracket appears outside the bracket, ie:

    ∫ f'(x) * (f(x))^n dx

    in this case f(x)=4x-1 and f'(x)=4. The mulitple in this case is 0.25 which you can take out of the integral and put it back in at the end to get the right answer:

    Code (Text):

    1 * (4x-1)^1.5 + c = (√(4x-1))^3 + c
    -   ----------       ------------
    4      1.5                6
     
     
  10. Oct 22, 2003 #9
    Lavalamp,
    That's how I learned to do it, way back in '87. I think it's much quicker and easier.
    Aaron
     
  11. Oct 22, 2003 #10
    I only learnt about integration at the end of 2002, and then I only learnt about sight integrals earlier this year. That's why I messed up the integration the first time round.
     
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