Question

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

Answer 1

`1/2`

Explanation:

First, find the bounds given by `f(x) = g(x)`

`(x-1)^3 = x-1`

`(x-1)^3 - (x-1) = 0`

`(x-1)((x-1)^2-1) = 0`

`(x-1)(x^2-2x) = 0`

`x(x-1)(x-2) = 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 `(x-1)^3 > x-1`, so in order to find the POSITIVE area between the two curves, we will subtract `x-1` (smaller) from `(x-1)^3` (bigger).

`int_0^1[(x-1)^3 - (x-1)]dx`

`int_0^1[x^3-3x^2+3x-1-x+1]dx`

`int_0^1(x^3-3x^2+2x)dx`

`[x^4/4-x^3+x^2]_0^1 = (1/4-1+1)-(0/4-0+0) = 1/4`

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

`int_1^2[(x-1)-(x-1)^3]dx`

`int_1^2[x-1-x^3+3x^2-3x+1]dx`

`int_1^2(-x^3+3x^2-2x)dx`

`[-x^4/4+x^3-x^2]_1^2 = (-16/4+8-4) - (-1/4+1-1) = 1/4`

So the area of the first section is `1/4` and the area of the second section is `1/4`. Therefore, the total area between `f(x)=(x-1)^3` and `g(x) = x-1` is `1/2`

Final Answer