Question

How do you find the area between `f(y)=y(2-y), g(y)=-y`?

Answer 1

`9/2 units^2`

Explanation:

Claculating the intersection Points:
`x(2-x)=-x`
so we get
`x(3-x)=0`
So we have
`x_1=0`
or
`x_2=3`
and we have
`int_0^3(x(2-x)+x)dx=int_0^3(3x-x^2)dx=9/2`

Answer 2

`A=int_0^3(3y-y^2)*dy=[3/2y^2-1/3y^3]_0^3=9/2`

Explanation:

The area between two curves due to y-axis is given by :

`color(red)[A=int_a^bx_2-x_1*dy`

`x_2=2y-y^2`

`x_1=-y`

lets find the cross between the curves:

`2y-y^2=-y rArr y^2-3y=0 rArr y(y-3)=0`

`y=3 or y=0`

`A=int_0^3(2y-y^2)-(-y)*dy`

`A=int_0^3(3y-y^2)*dy=[3/2y^2-1/3y^3]_0^3=9/2`

show the wanted area below (shaded):

`x_2=2y-y^2` red curve(green)

`x_1=-y` blue curve