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Math - Calculus III

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Find the double integral of f (x, y) = (x^7)y over the region between the curves y = x^2 and y = x(3 - x).

  • Math - Calculus III -

    The curves intersect at (0,0) and (3/2,9/4)

    So, we want

    Int[0,3/2] Int[x^2,3x-x^2] (x^7 y) dy dx

    = Int[0,3/2] (x^7 y^2)/2 [x^2,3x-x^2] dx

    = Int[0,3/2] (1/2)(9x^9 - 6x^10) dx

    = 1/2(9/10 x^10 - 6/11 x^11)[0,3/2]

    = 1/2 (3/2)^10 (9/10 - 6/11 * 3/2)

    = 1/2 (3/2)^10 (9/110)

    = 531441/225280 = 2.359

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