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Math Help - The integral....

  1. #1
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    The integral....

    Hello,
    plz try to solve that question.
    Thanks.
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    Quote Originally Posted by m777 View Post
    Hello,
    plz try to solve that question.
    Thanks.
    Do the inside integral, aka the y integral, first:
    Int[(4x + 2)dy, x^2, 2x]

    = [(4x + 2)y, x^2, 2x] <-- x is a constant for the y integral

    = (4x + 2)*2x - (4x + 2)*x^2

    = 8x^2 + 4x - 4x^3 - 2x^2

    = -4x^3 + 6x^2 + 4x

    Now do the outside integral, aka the x integral:
    Int[(-4x^3 + 6x^2 + 4x)dx, 0, 2]

    = [-4*(1/4)x^4 + 6*(1/3)x^3 + 4*(1/2)x^2, 0, 2]

    = -2^4 + 2*2^3 + 2*2^2

    = -16 + 16 + 8

    = 8

    -Dan
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