Posted by **Saima ** on Monday, March 1, 2010 at 7:27am.

Find the area of the region enclosed:

Y=x3+5x2+3, y=x2+7x +3,

X=0,x=3

- math -
**Reiny**, Monday, March 1, 2010 at 8:42am
We must first find where the two curves intersect.

x^3 + 5x^2 + 3 = x^2 + 7x + 3

x^3 + 4x^2 - 7x = 0

x(x^2 + 4x - 7)=0

so x=0 or x = -2 ± √11

so there is an intersection between our domain of x=0 to x=3 , this makes it quite messy.

We have to split the area into two regions

-one from 0 to -2+√11

-another from -2+√11 to 3

In the first region, the parabola is above the cubic, so the effective height is (x^2+7x+3) - (x^3 + 5x^2 + 3)

= -x^3 - 4x^2 + 7x

while for the second region the height will be

x^3 + 4x^2 - 7x

So Area

= integral[-x^3 - 4x^2 + 7x] from 0 to -2+√11 + integral[x^3 + 4x^2 - 7x] from -2+√11 to 3

= [(-1/4)x^4 - (4/3)x^3 + (7/2)x ] from 0 to -2+√11 + [(1/4)x^4 + (4/3)x^3 - (7/2)x ]from -2+√11 to 3

I will let you do the messy arithmetic,

Good Luck.

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