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Question regarding integration

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the integral:

    from 0 to pi/2
    I=[itex]\int[/itex]sin x {2f′(cos x) − 1} dx

    When f(0) = 2 and f(1) = 4

    2. Relevant equations

    3. The attempt at a solution

    integral from zero to pi/2 of 2[itex]\int[/itex]sin(x){f'(Cos (x) -1}dx

    u = cos (x)
    du = -sinx dx

    (Here is where I run into problems....i do not know know to do with the -1)

    1. We insert the limits of integration into the U = cos (x) and get

    0 = cos (pi/2)
    1 = cos (1)

    so now write

    -2[itex]\int[/itex]f'(u)-1du (limits of integration are from 1 to zero

    flip limits of integration
    2[itex]\int[/itex] f'u-1du (limits of integration are zero to 1)


    2(F(1)-1) - (2(F(0)-0)
    2(4-1) - 2(2-0)
    6 - 4 = 2.

    I tried 2 in my HM system and it is incorrect. What am I doing wrong?
  2. jcsd
  3. Sep 12, 2011 #2


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    Homework Helper

    OK, when you factored out the '2', you forgot to factor it out of the second term of 1 ....
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