# How do you find the area between f(x)=(x-1)^3 and g(x)=x-1?

Jun 1, 2017

$\frac{1}{2}$

#### Explanation:

First, find the bounds given by $f \left(x\right) = g \left(x\right)$

${\left(x - 1\right)}^{3} = x - 1$

${\left(x - 1\right)}^{3} - \left(x - 1\right) = 0$

$\left(x - 1\right) \left({\left(x - 1\right)}^{2} - 1\right) = 0$

$\left(x - 1\right) \left({x}^{2} - 2 x\right) = 0$

$x \left(x - 1\right) \left(x - 2\right) = 0$

$x = 0 , 1 , 2$

This means the graphs actually close off 2 separate areas, since there are three intersection points. To see what I mean by this, here is a graph of color(blue)(y=x-1 and color(red)(y = (x-1)^3:

So, to find the total area, we need to find the area of both sections and then add them together.

From $x = 0$ to $x = 1$, we can see that ${\left(x - 1\right)}^{3} > x - 1$, so in order to find the POSITIVE area between the two curves, we will subtract $x - 1$ (smaller) from ${\left(x - 1\right)}^{3}$ (bigger).

${\int}_{0}^{1} \left[{\left(x - 1\right)}^{3} - \left(x - 1\right)\right] \mathrm{dx}$

${\int}_{0}^{1} \left[{x}^{3} - 3 {x}^{2} + 3 x - 1 - x + 1\right] \mathrm{dx}$

${\int}_{0}^{1} \left({x}^{3} - 3 {x}^{2} + 2 x\right) \mathrm{dx}$

${\left[{x}^{4} / 4 - {x}^{3} + {x}^{2}\right]}_{0}^{1} = \left(\frac{1}{4} - 1 + 1\right) - \left(\frac{0}{4} - 0 + 0\right) = \frac{1}{4}$

From $x = 1$ to $x = 2$, we can see that $x - 1 > {\left(x - 1\right)}^{3}$, so in order to find the POSITIVE area between the two curves, we will subtract ${\left(x - 1\right)}^{3}$ (smaller) from $x - 1$ (bigger).

${\int}_{1}^{2} \left[\left(x - 1\right) - {\left(x - 1\right)}^{3}\right] \mathrm{dx}$

${\int}_{1}^{2} \left[x - 1 - {x}^{3} + 3 {x}^{2} - 3 x + 1\right] \mathrm{dx}$

${\int}_{1}^{2} \left(- {x}^{3} + 3 {x}^{2} - 2 x\right) \mathrm{dx}$

${\left[- {x}^{4} / 4 + {x}^{3} - {x}^{2}\right]}_{1}^{2} = \left(- \frac{16}{4} + 8 - 4\right) - \left(- \frac{1}{4} + 1 - 1\right) = \frac{1}{4}$

So the area of the first section is $\frac{1}{4}$ and the area of the second section is $\frac{1}{4}$. Therefore, the total area between $f \left(x\right) = {\left(x - 1\right)}^{3}$ and $g \left(x\right) = x - 1$ is $\frac{1}{2}$