mathematics

∫ _0^(1/2)xdx/√(1-x²)

the half should be at the top of this ∫ and the zero at the bottom.

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asked by lisa
  1. Ah, no problem

    We had for the indefinate integral

    Integral = -sqrt(1-x^2)
    so we need
    -sqrt(1-x^2) at x = 1/2
    minus
    -sqrt(1-x^2) at x = 0
    or
    -sqrt (1-1/4) + sqrt(1)
    =1 - sqrt(3/4)
    = 1 -(1/2) sqrt(3)

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    posted by Damon
  2. or we could have done it in t

    integral = -cos t
    where x = sin t
    when x = 0, t = 0 and cos t = 1
    when x = 1/2 , t = 30 degrees or pi/6
    then cos t = (1/2)sqrt 3
    so
    (1/2) sqrt(3) - 1

    sign difference is due to square root could be + or -

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    posted by Damon
  3. sign ambiguities
    -cos pi/6 = -(1/2)sqrt(3)

    -(-cos 0) = +1
    so
    - (1/2) sqrt(3) -1

    when sin t = 0, t could be 0 or 180 (pi)
    so cos t could be +1 or -1
    when sin t = 1/2
    t could be pi/6 or 5 pi/6
    and cos t could be -(1/2) sqrt(3)
    so that sqrt sign makes the signs of the answer tricky

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    posted by Damon

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