Calculus

The area bounded between the line y=x+4 and the quadratic function y=(x^2)-2x.
Hint: Draw the region and find the intersection of the two graphs. Add and subtract areas until the appropriate area is found.

I found the intersection points as (-1,3) and (4,8).
I'm not sure what to do with this. I think the answer is 125/6, but not sure.
Please show me the steps on how to get this answer if it's correct.

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asked by Janet
  1. Your intersection points are correct. Between those points, the y = x+4 curve is above the y = x^2 -2x curve.

    The area between the curves is the integral of x + 4 dx from x = -1 to 4, MINUS the integral of x^2 -2x between the same two x values. That equals

    x^2/2 + 4x @ x = 4 - (x^2/2 +4x) @ x = -1
    MINUS
    x^3/3 -x^2 @ x = 4 - (x^3/3 -x^2) @x = -1

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    posted by drwls
  2. Your intersections are correct.
    the vertex of the parabola is at (1,-1)
    The zeros of the parabola are at(0,0) and at (2,0)
    Between x = 0 and x = 2, the parabola dips below the x axis

    we want the height of the line between y = x+4 and y=x^2-2x
    which is
    x+4 -x^2+2x
    or
    -x^2 + 3x + 4
    integrate that dx from -1 to +4

    -x^3/3 + 3x^2/2 + 4x

    at 4
    -64/3 + 24 + 16 = (-64+72+48)/3
    = 56/3

    at -1
    +1/3 +3/2 -4 = 2/6 + 9/6 - 24/6
    = -13/6
    so we want
    56/3 -(-13/6) = 112/6+13/6 = 125/6
    yep, agree 125/6

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    posted by Damon
  3. Yeah. Thanks for showing me the steps

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    posted by Janet

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