Posted by Jose on .
Find the area of the region bounded by y=x^2 and y = (x4)^2 +4 and the lines y=0 and y=4.

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

Calc 
Damon,
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 (y10) dx
where y1 = x^2
from x = 0 to x = 4
dx x^2 = x^3/3 = 64/3
now subtract integral of dx[(x4)^2 + 4] from x = 2 to x = 4
dx[x^28 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 !!! 
Calc 
Reiny,
I made a sketch, which showed the 2 parabolas don't intersect.
so you have the open space between them cut off by the xaxis (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 xaxis 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[(x4)^2 + 4] from x = 2 to x = 4 >>