find the area of the region enclosed by the given curves. y=x^4 +4x and y=3x^2 +4x+4

Answer 1

Let z=x^2

y=z^2 + 4x = 3z + 4x + 4

4x cancels

leaving z^2 - 3z + 4 = 0

that factors

(z-4)(z+1) = 0

z = -1 or 4, x= sqr4 = + or - 2

those are the two places the curves intersect

substitute x=0 into the 2 equations and in the interval (-2,2) and the quadratic equation is on top

integrate the quadratic minus the 4th power equation on the limits -2 to 2 to get the area between them

x^5 - x^3 + 4x evaluated at 2 and -2 gives 64/5 as the area or 12 4/5