Question

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

Answer 1

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.