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.

Answer 1

i got 16, but i am not sure if i am right

Answer 2

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 !!!

Answer 3

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 >>