Question

Finding exact area of region by intergration?

Sketch the graph of `y=x(2-x)(x-1)` and find the exact area of the region enclosed by the curve and the x-axis.

Answer 1

I tried this:

Explanation:

I used Excel to plot the graph:

The two areas of interest are the green ones in the above picture going from `x=0` to `x=1` and from `x=1` and `x=2`.
We can write:

`A=int_a^b[x(2-x)(x-1)]dx=int_a^b(3x^2-2x-x^3)dx=3x^3/3-2x^2/2-x^4/4=x^3-x^2-x^4/4`
evaluated between our points of interest.

Let us then consider the two areas separately:

`A_1=int_0^1(3x^2-2x-x^3)dx=(1-1-1/4)-0=-1/4"u.a."`
the negative sign arises because the area lays below the `x` axis. We can consider it as a positive value giving a contribution of `A_1=1/4"u.a."`
and

`A_2=int_1^2(3x^2-2x-x^3)dx=(8-4-16/4)-(1-1-1/4)=1/4"u.a."`

The total area will be:

`A=A_1+A_2=1/4+1/4=2/4=1/2 "u.a."`