Calculus Area Cubic curve

Question

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?

Answer 1

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}

Answer 2

Given is

$$ f(x) = x^3 $$

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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 $B$.

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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 $C$.

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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

$$ \frac{ \int_{-2}^{+4} \Big( - \xi^3 + 12 \xi + 16 \Big) d \xi } { \int_{-2}^{+1} \Big( \xi^3 - 3 \xi + 2 \Big) d \xi } = \frac{108}{27/4} = 16 $$

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So the ratio of these surfaces is

$$ 16 $$

for any point $A$.