Question

how do you find the area between x/3=y and x=y^2-4
https://s3.amazonaws.com/grapher/exports/ox7nfn4kf5.png
ans:125/6

Answer 1

by integration or without use integration ?

Answer 2

doesn't matter.

Answer 3

Integration would be easier I assume

Answer 4

easier using the formula :
area = D * sqrt(D)/6a^2
with D is the value of discriminant of quadratic equation : ax^2+bx+c=0

x/3=y ------> x=3y
x=y^2-4

x1=x2
y^2-4=3y
y^2-3y-4=0
D=(-3)^2-4(1)(-4) = 9+16=25
area = Dsqrt(D)/6a^2 = 25sqrt(25)/6(1)^2=125/6