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calculus

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find the area of y=x sqrt(x^2+16), bound by the x-axis and the vertical line x=3
I got 22.5 is that correct?

  • calculus - ,

    The function crosses the origin
    so I see it as

    A = ∫ x(x^2+16)^(1/2) dx from 0 to 3
    = [ (1/3)(x^2+16)^(3/2) ] from 0 to 3
    = (1/3)(25)^(3/2) - 0
    = (1/3)(125)
    = 125/3

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