`y = x^3, y = x` Sketch the region enclosed by the given curves and find its area.

You need to determine first the points of intersection between curves `y = x^3 ` and `y = x` , by solving the equation, such that:

`x^3 = x => x^3 - x = 0`

Factoring out x yields:

`x(x^2 - 1) = 0 => x = 0` or` x^2 - 1 = 0 => x = 1` and `x = -1`

Hence, the endpoints of integral are x = -1, x = 0, x...

You need to determine first the points of intersection between curves `y = x^3 ` and `y = x` , by solving the equation, such that:

`x^3 = x => x^3 - x = 0`

Factoring out x yields:

`x(x^2 - 1) = 0 => x = 0` or` x^2 - 1 = 0 => x = 1` and `x = -1`

Hence, the endpoints of integral are x = -1, x = 0, x = 1.

You need to decide what curve is greater than the other on the interval [-1,1]. You need to notice that `x^3>x` on the interval [-1,0], and` x^3<x` on [0,1] hence, you may evaluate the area of the region enclosed by the given curves, such that:

`int_a^b (f(x) - g(x))dx,` where f(x) > g(x) for `x in [a,b]`

`int_(-1)^1 |x^3 - x|dx = int_(-1)^0 (x^3 - x) dx + int_0^1 (x - x^3) dx`

`int_(-1)^1 |x^3 - x|dx = int_(-1)^0 x^3 dx - int_(-1)^0 xdx + int_0^1 x dx - int_0^1 x^3 dx`

`int_(-1)^1 |x^3 - x|dx = x^4/4|_(-1)^0 - x^2/2_(-1)^0 + x^2/2|_0^1 - x^4/4|_0^1`

`int_(-1)^1 |x^3 - x|dx = 0 - 1/4 - 0 + 1/2 + 1/2 - 0 - 1/4 + 0`

`int_(-1)^1 |x^3 - x|dx = 1 - 1/2`

`int_(-1)^1 |x^3 - x|dx = 1/2`

Hence, evaluating the area of the region enclosed by the given curves, yields `int_(-1)^1 |x^3 - x|dx = 1/2.`

The area of the region enclosed by the given curves is found between the red and orange curves, for `x in [-1,1].`