HELP!!!!!! Math Geniuses

The run-off next to a race track is described by the area enclosed between the curve y=x(x-10)^2+2x^4 and the curve y=2x^4 where x and y are measured in metres. Find the area of the run-off, establishing first where the curves intersect.

1- to find where the curves intersect we solve the equation x(x-10)^2+2x^4 = 2x^4 which gives two roots x=10 and x=0 , thus the area is enclosed between these two points of intersection
2- to find the enclosed area you have to integrate the function y= (x(x-10)^2+2x^4) - ( 2x^4) = x(x-10)^2 from x=0 to x=10 which gives x^4/4 - 20x^3/3 + 100x^2/2 where x=10
3- enclosed area = 10000/4 -20000/3 + 10000/2 = 2500 - 6666.67 + 5000 = 833.33

First solve simultaneously to find the x-coordinate of their point of intersection. Since y=x(x-10)^2+2x^4 and y=2x^4, then, x(x-10)^2+2x^4=2x^4, get x=0 or x=10. The required area, A=int(x(x-10)^2+2x^4-2x^4) from x=0 to x=10. Thus, A=int(x(x-10)^2). Integration by parts gives A=10(x-10)^3, between x=0 and x=10, thus, A=10(10-10)^3-10(0-10)^3=10000m^2

Curves intersect where y = y
2x^4=2x^4+x(x-10)^2, which turns to
0=x(x-10)^2, x is thus equal to 0 and 10. Now you need to find the upper bound and lower bound so pick a number between one and ten then plug it in, like one. So in Ur first equation when x =1, y = 83, and in the other y =2. so your first equation is the upper bound. Thus integrate from one to ten your first equation minus your second

they intersect where the point (x, y) is verified by both functions

y1=y2
x(x - 10)² + 2x⁴ = 2x⁴

x(x - 10)² = 0

x = 0
x = 10

A = |∫(from 0 to 10)[x(x - 10)² + 2x⁴- 2x⁴]| =

= |∫ (x³ -20 x² + 100x)| =

=| (x⁴)/4 - 20/3x³ + 50x²| [from 0 to 10] =

= |2500 - 6666.666 + 5000| = 833.333