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calculus

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find the area of the region enclosed by y=squareroot of x, the line tangent to y= squareroot of x at x=4 and the x-axis

  • calculus -

    y = √x
    y' = 1/2√x

    So, at (4,2) the tangent line has slope 1/4.

    So, the equation for the line is
    (y-2)/(x-4) = 1/4
    y = x/4 + 1

    which insersects tha x-axis at x = -4.

    Now, √x is not defined for x<0, so we need to break the region into two parts:

    Int(x/4 + 1)dx [-4,0] + Int(x/4 + 1 - √x)dx [0,4]

    = [x^2/8 + x)[-4,0] + (x^2/8 + x - 2/3x^3/2)[0,4]

    8 - 16/3 = 8/3

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