Posted by Janet on Monday, May 9, 2011 at 10:00pm.
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.

Calculus  drwls, Monday, May 9, 2011 at 10:18pm
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

Calculus  Damon, Monday, May 9, 2011 at 10:36pm
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^22x
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

Calculus  Janet, Monday, May 9, 2011 at 10:49pm
Yeah. Thanks for showing me the steps
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