Calculus Area Cubic curve Why area bounded between the "line $AB$" and the "cubic curve" and area bounded between the "line $BC$" and the "cubic curve" is $16$ times? 

 A: The following information can be extracted from the picture
\begin{align}
y_{purple}(x) &= 972x+11664 \\
y_{blue}(x) &= 243x-1458 \\
y_{green}(x) &=x^3
\end{align}
We also need the intersection points
\begin{align}
A &=(9,729) \\
B &=(-18,-5832) \\
C &=(36,46656)
\end{align}
The purple area can be computed as
\begin{align}
A_{purple} &= \int_{-18}^{36} y_{purple}(x)-y_{green}(x)\,dx \\
&= \int_{-18}^{36} 972x+11664-x^3\,dx \\
&= 708588
\end{align}
And the green area as
\begin{align}
A_{greem} &= \int_{-18}^{9} y_{green}(x)-y_{blue}(x)\,dx \\
&= \int_{-18}^{9}x^3-243x+1458\,dx \\
&= \frac{177147}{4}
\end{align}
If we compare the areas, we get
\begin{align}
\frac{A_{purple}}{A_{green}} = \frac{708588}{\frac{177147}{4}}=16
\end{align}
A: Given is
$$
f(x) = x^3
$$

For the point $A$, $x=x_o$, we have a tangent
$$
3 x_o^2
$$
So that line is given by
$$
y = \Big(3 x - 2 x_o \Big) x_o^2
$$
The intersection is given by the equation
$$
x^3 - 3 x x_o^2 + 2 x_o^3 = 0
$$
which can be written as
$$
\Big( x - x_o \Big)^2 \Big( x + 2 x_o \Big) = 0
$$
As for point $A$ we have $x = x_o$, we get $x = - 2 x_o$ for point $B$.

For the point $B$, $x =-2 x_o$, we have a tangent
$$
12 x_o^2
$$
So that line is given by
$$
y = 4 \Big(3 x - 4 x_o \Big) x_o^2
$$
The intersection is given by the equation
$$
x^3 - 12 x x_o^2 - 16 x_o^3 = 0
$$
which can be written as
$$
\Big( x + 2 x_o \Big)^2 \Big( x - 4 x_o \Big) = 0
$$
As for point $B$ we have $x = - 2x_o$, we get $x = 4 x_o$ for point $C$.

We now have two surfaces
$$
S_{BA} = \int_{-2x_o}^{+x_o} \Big( x^3 - 3 x x_o^2 + 2 x_o^3 \Big) d x
= x_o^2 \int_{-2}^{+1} \Big( \xi^3 - 3 \xi + 2 \Big) d \xi
$$
and
$$
S_{BC} = \int_{-2x_o}^{+4x_o} \Big( - x^3 + 12 x x_o^2 + 16 x_o^3 \Big) d x
= x_o^2 \int_{-2}^{+4} \Big( - \xi^3 + 12 \xi + 16 \Big) d \xi
$$
The ratio of these surfaces is constant and is given by
$$
\frac{ \int_{-2}^{+4} \Big( - \xi^3 + 12 \xi + 16 \Big) d \xi }
{ \int_{-2}^{+1} \Big( \xi^3 - 3 \xi + 2 \Big) d \xi } = \frac{108}{27/4} = 16
$$

So the ratio of these surfaces is
$$
16
$$
for any point $A$.
