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Solve the definite integral?

  1. Dec 7, 2009 #1
    The problem is: ∫ x / sqrt(x²+1)

    Upper limit is sqrt 3
    lower limit is 0

    I choose u as x²+1

    1/2du = dx.x

    1/2∫ 1/u^1/2 =

    1/2∫ u^(-1/2)du =

    1/2 ∫ U^(1/2) / (1/2) =

    (1/2) * (2/1) * u^1/2 = 1 * u^(1/2)


    u = (sqrt3)² + 1 = 4
    u = 0² + 1 = 1

    than i substituted

    1 * 4^1/2 - 1 * 1^(1/2) =
    1 * 2 - 1 * 1 =
    2 - 1 = 1

    is this correct?
     
    Last edited: Dec 7, 2009
  2. jcsd
  3. Dec 7, 2009 #2

    Mark44

    Staff: Mentor

    Yes. Here's another way that changes the limits of integration for the same substitution.

    [tex]\int_0^{\sqrt{3}} \frac{x~dx}{\sqrt{x^2 + 1}}[/tex]
    [tex]= \frac{1}{2}\int_1^4 u^{-1/2}~du~=~u^{1/2}|_1^4[/tex]
    [tex]=2 - 1 = 1[/tex]
     
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