Find the area enclosed by the curves y 6x and y x2Solution

Solution

, first off you need to find the points where these two graphs meet. so what you do is
x^2-6x=-x^2+4x
2x^2-10x=0

2x(x-5)=0

x1=0
x2=5

okay, so now we make a specific integral from 5 to 0 so:

int(-x^2+4x(the highest graph) -(x^2-6x)(the lowest graph) dx, 0,5)

int(-2x^2+10x,dx,0,5)

int(-(2x^3)/3+5x^2,dx,0,5)

int(-(2*5^3)/3+5*5^2 - 0) = 41.67