You need to find first where the curves intersect each other, hence, you need to solve the equation `f(x) = g(x)` such that:

`x^3-6x^2+8x = x^2-4x`

Moving all terms to one side yields:

`x^3 - 6x^2 + 8x - x^2 + 4x = 0`

`x^3 - 7x^2 + 12x = 0`

Factoring out x yields:

`x(x^2 - 7x + 12) = 0 => x = 0`

`x^2 - 7x + 12 = 0 `

You need to use quadratic formula such that:

`x_(1,2) = (7+-sqrt(49 - 48))/2 => x_(1,2) = (7+-1)/2`

`x_1 = 4 ; x_2 = 3`

You need to check what curve is above over the intervals [0,3] and [3,4].

`If x = 1 => f(1) = 1-6+8 = 3`

`g(1) = 1-4 = -3`

Hence `f(1)>g(1), ` thus, the curve of the function f(x) is above the curve of the function g(x), over the interval `[0,3].`

`f(3.5) = 42.875- 73.5 + 28 = -2.625`

`g(3.5) = 12.25 - 14 = -1.75`

Notice that `g(x)>f(x)` over the interval `[3,4].`

Hence, evaluating the area bounded by the given curves yields:

`A = int_0^3 (f(x) - g(x))dx + int_3^4(g(x - f(x))) dx`

`A = int_0^3 (x^3 - 7x^2 + 12x) dx + int_3^4(7x^2 - 12x - x^3) dx`

Using the fundamental theorem of calculus yields:

`A = (x^4/4 - 7x^3/3 + 12x^2/2)|_0^3 + (7x^3/3 - 12x^2/2 - x^4/4)|_3^4`

`A = (81/4 - 63 + 54) + (448/3 - 96 - 64 - 63 + 54 + 81/4)`

`A = 81/2 - 126 + 108 - 160 + 448/3`

`A = 1139/6 - 178 => A = (1139 - 1068)/6 => A = 71/6`

**Hence, evaluating the area of the region bounded bby the given curves yields `A = 71/6.` **