Posted by Saima on .
Find the area of the region enclosed:
Y=x3+5x2+3, y=x2+7x +3,
X=0,x=3

math 
Reiny,
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.