Find the area of the region bounded by y=x^2 and y = -(x-4)^2 +4 and the lines y=0 and y=4.
i got 16, but i am not sure if i am right
Sketch it.
parabola open on top with vertex at (0,0) then (2,4), (4,16)
parabola open on bottom with vertex at (4,4) and crossing the x axis at x = 2 and x = 6
do integral of (y1-0) dx
where y1 = x^2
from x = 0 to x = 4
dx x^2 = x^3/3 = 64/3
now subtract integral of dx[(x-4)^2 + 4] from x = 2 to x = 4
dx[x^2-8 x + 20]=x^3/3 -4x^2 + 20 x
at 4 = 64/3 -64 + 80
at 2 = 8/3 - 16 + 40
difference = 56/3 - 48 + 20 = 56/3 -28
so subtract
64/3 - 56/3 + 28
8/3 + 28
92/3
check my arithmetic !!!
I made a sketch, which showed the 2 parabolas don't intersect.
so you have the open space between them cut off by the x-axis (y=0) and the horizontal y = 4
A vertical form (2,0), on the second
and (2,4), which lies on the first,
splits the area into two equal regions.
There is symmetry, since the two parabolas are congruent
so all we need is double the area between y = x^2 and the x-axis from 0 to 2, which is
2[integral] x^2 dx from 0 to 2
= 2 (x^3/3 │ from 0 to 2)
= 2(8/3) = 16/3
I then did it the long way and found the area in two parts
area = [integral] x^2 dx from 0 to 2
+ [integral] (4 - (-(x+4)^2 + 4)) dx from 2 to 4
and got 8/3 + 8/3 = 16/3
BTW, I did notice that Damon missed the negative sign in front of the (x+4)^2 + 4 in his line
<< now subtract integral of dx[(x-4)^2 + 4] from x = 2 to x = 4 >>
To find the area of the region bounded by the given curves and lines, we need to first determine the x-coordinates of the points where these curves intersect.
We begin by setting the two equations for the curves equal to each other:
x^2 = -(x-4)^2 + 4
Expanding the right side of the equation:
x^2 = -(x^2 - 8x + 16) + 4
x^2 = -x^2 + 8x - 16 + 4
2x^2 - 8x + 12 = 0
Now, we can solve this quadratic equation for x using factoring or the quadratic formula. In this case, factoring is the most convenient method, so let's factor the equation:
2(x^2 - 4x + 6) = 0
Setting each factor equal to zero:
x^2 - 4x + 6 = 0
Unfortunately, this quadratic equation can't be factored further, so we proceed to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -4, and c = 6. Substituting these values into the formula:
x = (-(-4) ± √((-4)^2 - 4(1)(6))) / (2(1))
x = (4 ± √(16 - 24)) / 2
x = (4 ± √(-8)) / 2
Since we have a negative value inside the square root, it means that the given equations do not intersect. Therefore, there is no region enclosed, and the area is equal to zero.