We will have the following:
[tex]A=\int ^1_0(12y^2-12y^3)\partial y-\int ^1_0(2y^2-2y)\partial y\Rightarrow A=\int ^1_0(12y^2)\partial y-\int ^1_0(12y^3)\partial y-\int ^1_0(2y^2)\partial y+\int ^1_0(2y)\partial y[/tex]
Here we remember that the antiderivative follows:
[tex]\int abn^{b-1}\partial n=an^b+c[/tex]
So, we will have:
[tex]A=(4y^3|^1_0-(3y^4|^1_0-(\frac{2}{3}y^3|^1_0+(y^2|^1_0[/tex][tex]\Rightarrow A=(4(1)^3-4(0)^3)-(3(1)^4-3(0)^4)-(\frac{2}{3}(1)^3-\frac{2}{3}(0)^3)+(1^2-0^2)[/tex][tex]\Rightarrow A=4-3-\frac{2}{3}+1\Rightarrow A=\frac{4}{3}[/tex]
So, the area between the two lines is of 4/3 square units.
