find the region enclosed by the given curves:
4x+y^2=9, x=2y
I assume you want the area of the region
you will need their intersection
use substitution ....
4(2y) + y^2 = 9
y^2 + 8y - 9 = 0
(y-1)(y+9) = 0
y = 1 or y -9
if y = 1 , x = 2 --> (2,1)
if y = -9, x = -18 --> (-18, -9)
use horizontal slices
x = -(1/4)y^2 + 9/2
the value of x for the region
= (-1//4)y^2 + 9/2 - 2y
area = ∫(-1/4 y^2 + 9/2 - 2y ) dy from -9 to 1
= [(-1/20)y^5 + (9/2)y - y^2 ] from -9 to 1
= (-1/20) + 9/2 - 1 - ( (-1/20)(-59049) + 81/2 - 18)
= ....
you do the button - pushing.
(you might also want to check my arithmetic, my weakness)