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
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.
3 answers
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
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
Yeah. Thanks for showing me the steps
1 answer