Posted by **Jose** on Thursday, April 2, 2009 at 1:10am.

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.

- Calc -
**Jose**, Thursday, April 2, 2009 at 1:38am
i got 16, but i am not sure if i am right

- Calc -
**Damon**, Thursday, April 2, 2009 at 2:00am
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 !!!

- Calc -
**Reiny**, Thursday, April 2, 2009 at 9:14am
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 >>

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