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February 27, 2015

February 27, 2015

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

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

- Calc -
**Damon**, Thursday, April 2, 2009 at 2:00amSketch 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 9:14amI 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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