Question

The regions bounded by the graphs of y=4x and y=x−x2+x3 are shaded in the figure above. What is the sum of the areas of the shaded regions?

Answer 1

Find where the two curves intersect:

$4x = x - x^2 + x^3$

$\implies x^3 - x^2 - 3x = 0$

$\implies x (x^2 - x - 3) = 0$

$\implies x = 0 \text{ or } x^2 - x - 3 = 0$

Completing the square gives

$x^2 - x - 3 = \left(x - \dfrac12\right)^2 - \dfrac{13}4 = 0$

and solving for x,

$\implies x = \dfrac12 \pm \sqrt{\dfrac{13}4} = \dfrac{1 \pm \sqrt{13}}2$

The total unsigned area between the two curves is then

$\displaystyle \int_{(1-\sqrt{13})/2}^{(1+\sqrt{13})/2} \left|4x - \left(x - x^2 + x^3\right)\right| \, dx$

$= \displaystyle -\int_{(1-\sqrt{13})/2}^0 \left(4x - \left(x - x^2 + x^3\right)\right) \, dx + \int_0^{(1+\sqrt{13})/2} \left(4x - \left(x - x^2 + x^3\right)\right) \, dx$

$= \displaystyle -\int_{(1-\sqrt{13})/2}^0 \left(3x + x^2 - x^3\right) \, dx + \int_0^{(1+\sqrt{13})/2} \left(3x + x^2 - x^3\right) \, dx$

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

$\displaystyle \int (3x + x^2 - x^3) \, dx = \dfrac{3x^2}2 - \dfrac{x^3}x - \dfrac{x^4}4 + C$

so that by the fundamental theorem of calculus, the area we want is

$\displaystyle \int_{(1-\sqrt{13})/2}^{(1+\sqrt{13})/2} \left|4x - \left(x - x^2 + x^3\right)\right| \, dx = \boxed{\dfrac{73}{12}}$