What is the area bounded by $$y=x^3-7x^2+14x-8$$ and the x-axis in the interval $$[0,2]$$?

Factor the polynomial to find where the curve meets the x axis:

$$y = x^3 - 7x^2 + 14x - 8$$

$$y = (x - 1)(x - 2)(x - 4)$$

The required interval goes from zero to two.

The roots of the polynomial inside this range are at $$x = 1$$ and $$x = 2$$.

The curve changes sign at $$x = 1$$:

• From zero to one the function is negative so we take the negative of the integral
  
• From one to two the function is positive so we take the standard integral
  

Set up the area calculation:

$$\text{Area} = -\int_0^1 (x^3 - 7x^2 + 14x - 8) \, dx + \int_1^2 (x^3 - 7x^2 + 14x - 8) \, dx$$

Find the antiderivative function:

$$F(x) = \frac{x^4}{4} - \frac{7x^3}{3} + 7x^2 - 8x$$

Evaluate the first integral from zero to one:

$$-\left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = -\left( \frac{3 - 28 - 12}{12} \right) = \frac{37}{12}$$

Evaluate the second integral from one to two:

$$\left( 4 - \frac{56}{3} + 28 - 16 \right) - \left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = \left( 16 - \frac{56}{3} \right) - \left( -\frac{37}{12} \right)$$

$$= -\frac{8}{3} + \frac{37}{12} = \frac{-32 + 37}{12} = \frac{5}{12}$$

Sum both portions to get the total area:

$$\text{Total Area} = \frac{37}{12} + \frac{5}{12} = \frac{42}{12} = \frac{7}{2}$$

The correct option is D.