Question

What is the area between the line `y = x` and the curve `y = x^3`?

Answer 1

Area `= 1/2 \ "unit"^2`

Explanation:

Graphs of the functions `y=x^3` and `y=x` are:
graph{(y-x^3)(y-x)=0 [-3.465, 3.464, -1.733, 1.73]}

Both `y=x^3` and `y=x` are odd functions so by symmetry we only need to consider the RHS and double it. We can see from the graph that the (positive) intersection points are `(0,0)` and `(1,1)` which we can verify algebraical:

`x^3=x => x^3-x=0 => x(x^2-1)=0 => x-0,+-1`

The area bounded by the `x`-axis, `y`-axis and `y=x` is that of a triangle with area:

`A_1 = 1/2(1)(1) = 1/2`

The area bounded by the `x`-axis, `y`-axis and `y=x^3` is given by:

`A_2 = int_0^1 \ x^3 \ dx`
`\ \ \ \ =[1/4x^4]_0^1`
`\ \ \ \ =1/4`

Then the total bounded area that we seek (LHS and RHS) is given by:

`A=2(A_1-A_2)`
`\ \ \ =2(1/2-1/4)`
`\ \ \ =1/2`

Answer 2

`1/2` unit`"s"^2`

Explanation:

Start by finding the points of intersection by solving the system `{(y = x^3), (y = x):}`

`x^3 = x`

`x^3 - x = 0`

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

`x(x + 1)(x - 1) = 0`

`x = 0, 1 and -1`

So, we have to find the area in the interval `0 ≤ x ≤ 1` and in the interval `-1 ≤ x ≤ 0`. If you trace the graphs of the two functions, you will find that the area in `-1 ≤ x ≤ 0` lies beneath the y-axis while the area in `0 ≤ x ≤1` will lie above the y-axis. We simply add the two areas up in this case.

Here is the graph:

As you can see, in the interval `0 ≤ x ≤ 1`, the line `y = x` lies above `y = x^3` and in the interval `-1 ≤ x ≤ 0`, `y = x^3` lies above `y = x`. Our integrals will therefore be `int_-1^0 (x^3 - x)` and `int_0^1 (x - x^3)`

Evaluating `int_-1^0 (x^3 - x) dx`

Integrate using the rule `int(x^n)dx = x^(n + 1)/(n + 1)`:

`= [1/4x^4 - 1/2x^2]_-1^0`

Evaluate using `int_a^bF(x)dx = f(b) - f(a)`, where `f'(x) = F(x)`.

`= 1/4(0)^4 - 1/2(0)^2 - (1/4(-1)^4 - 1/2(-1)^2)`

`=-1/4 + 1/2`

`= 1/4`

Evaluating `int_0^1x - x^3dx`

Use the same process as above:

`=[1/2x^2 - 1/4x^4]_0^1`

`=1/2(1)^2 - 1/4(1)^2 - (1/2(0)^2 - 1/4(0)^4)`

`= 1/4`

Add the two areas to find the total area.

`A_"total" = 1/4 + 1/4 = 1/2`

Hence, the area between `y= x^3` and `y = x` is `1/2` unit`"s"^2`.

Hopefully this helps!