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March 3, 2015

March 3, 2015

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

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 10:18pmYour 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 10:36pmYour 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

- Calculus -
**Janet**, Monday, May 9, 2011 at 10:49pmYeah. Thanks for showing me the steps

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