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^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 10:49pm
Yeah. Thanks for showing me the steps

## Answer this Question

## Related Questions

- Calculus - 1. Find the area of the region bounded by the curves and lines y=e^x ...
- Calculus-Steve - Find the area of the region between the graphs of f(x)=3x+8 and...
- calculus - 2. Sketch the region in the first quadrant that is bounded by the ...
- Calculus AB - Let R be the region bounded by the graphs of y=sin(pi x) and y=(x^...
- calculus - 1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x...
- Calculus - Let f be the function given by f(x)=(x^3)/4 - (x^2)/3 - x/2 + 3cosx. ...
- Calc - Find the area of the region bounded by the curves y2 = x, y – 4 = x, y...
- Calculus - The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3 A...
- calculus - Consider the graphs of y = 3x + c and y^2 = 6x, where c is a real ...
- calculus - 1. Let S be the region in the first quadrant bounded by the graphs of...

More Related Questions