Find where the two curves intersect:
[tex]4x = x - x^2 + x^3[/tex]
[tex]\implies x^3 - x^2 - 3x = 0[/tex]
[tex]\implies x (x^2 - x - 3) = 0[/tex]
[tex]\implies x = 0 \text{ or } x^2 - x - 3 = 0[/tex]
Completing the square gives
[tex]x^2 - x - 3 = \left(x - \dfrac12\right)^2 - \dfrac{13}4 = 0[/tex]
and solving for x,
[tex]\implies x = \dfrac12 \pm \sqrt{\dfrac{13}4} = \dfrac{1 \pm \sqrt{13}}2[/tex]
The total unsigned area between the two curves is then
[tex]\displaystyle \int_{(1-\sqrt{13})/2}^{(1+\sqrt{13})/2} \left|4x - \left(x - x^2 + x^3\right)\right| \, dx[/tex]
[tex]= \displaystyle -\int_{(1-\sqrt{13})/2}^0 \left(4x - \left(x - x^2 + x^3\right)\right) \, dx + \int_0^{(1+\sqrt{13})/2} \left(4x - \left(x - x^2 + x^3\right)\right) \, dx[/tex]
[tex]= \displaystyle -\int_{(1-\sqrt{13})/2}^0 \left(3x + x^2 - x^3\right) \, dx + \int_0^{(1+\sqrt{13})/2} \left(3x + x^2 - x^3\right) \, dx[/tex]
where we use the definition of the absolute value function to split up the integral at the point where one curve "overtakes" the other - in other words, y = 4x lies above the curve y = x - x² + x³ when x > 0, and vice versa otherwise.
We have the antiderivative
[tex]\displaystyle \int (3x + x^2 - x^3) \, dx = \dfrac{3x^2}2 - \dfrac{x^3}x - \dfrac{x^4}4 + C[/tex]
so that by the fundamental theorem of calculus, the area we want is
[tex]\displaystyle \int_{(1-\sqrt{13})/2}^{(1+\sqrt{13})/2} \left|4x - \left(x - x^2 + x^3\right)\right| \, dx = \boxed{\dfrac{73}{12}}[/tex]
