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Definite Integrals Calculus I

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_{-4}^{2} f(x) dx =9, \int_{-4}^{-2} f(x) dx=2, \int_{0}^{2} f(x)dx =3[/tex]

    Find (I have solved this one)
    [tex]\int_{-2}^{0} f(x)dx= 4[/tex]

    [tex]\int_{0}^{-2} (9 f(x)- 2)dx= ?[/tex]

    2. Relevant equations

    [tex]\int_{a}^{b} f(x) dx = F(b) - F(a)[/tex]

    3. The attempt at a solution

    I don't see where I am doing something wrong.

    [tex]\int_{0}^{-2} (9 f(x)- 2)dx[/tex]

    [tex]-\int_{-2}^{0} (9 f(x)- 2)dx[/tex]

    Using the fundamental theorem of calculus.

    -[(9b(4) - 2b) - (9a(4) - 2a)]
    -[(0(4) - 0) - (9(-2)(4) - 2(-2))]

    Solving for that I get


    Therfore I figure that

    [tex]\int_{0}^{-2} (9 f(x)- 2)dx= -68[/tex]

    However the electronic homework says it is incorrect, but why?
  2. jcsd
  3. Dec 5, 2009 #2


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    Homework Helper

    may be easier to see if you write it out as
    [tex]\int_{0}^{-2} (9 f(x)- 2)dx[/tex]

    [tex]=-\int_{-2}^{0} (9 f(x)- 2)dx[/tex]

    [tex]=-9\int_{-2}^{0} f(x)dx+\int_{-2}^{0} 2dx[/tex]
  4. Dec 5, 2009 #3
    Ahah, integration properties! Thank you.
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