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Thread: Integrals

  1. #1
    Feb 2009
    Hello again =/

    I had a question about using integral calculus. I need to find the area enclosed by the following function:

    f(x) = x(x-1)(x-3) and the lines x = 0 and x = 4

    I thought that since f(x) = x^3 - 4x^2 +3x F(x) would be (1/4x)^4 - (4/3)x^3 + (3/2)x^2. Knowing that the integral is positive on [0,1], negative on [1,3] and positive again on [3,4] I just filled everything in. (x(1)-x(0) - (x(3)-(x1) + (x(4)-(x(3)

    I filled everything in and found the solution 13/4, but according to the answers it must be 8 (doesn't say why though). Can anybody help me solve this problem or tell me where I'm wrong? Much appreciated!

    No Edit button to be found anywhere...

    Just wanted to say I solved it. Apparently, I'm just stupid and forgot a minus sign on the last equation (x(4)-(x(3) which made the outcome indeed 8 =/

    ~Can be closed asap~
    Last edited by Krizalid; March 4th 2009 at 10:25 AM.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Mar 2007
    Santiago, Chile
    I just merged your posts, but, it's not appropriate to delete this 'cause may help anyone, of course, by providing your solution.
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