1. **State the problem:** Find the total area of the shaded regions bounded by the curve $f(x) = x^3 - 6x^2 + 8x$ and the x-axis between $x=0$ and $x=4$, where the curve crosses the x-axis at points $(0,0)$, $(2,0)$, and $(4,0)$. 2. **Formula and rules:** The area between a curve and the x-axis from $a$ to $b$ is given by the definite integral $$\text{Area} = \int_a^b |f(x)| \, dx$$ Since the function crosses the x-axis, the function changes sign, so we must split the integral at the roots and take the absolute value of each integral. 3. **Find the sign of $f(x)$ on each interval:** - For $0 < x < 2$, test $x=1$: $f(1) = 1 - 6 + 8 = 3 > 0$ (positive) - For $2 < x < 4$, test $x=3$: $f(3) = 27 - 54 + 24 = -3 < 0$ (negative) 4. **Calculate the area:** Split the integral into two parts: $$\text{Area} = \int_0^2 f(x) \, dx - \int_2^4 f(x) \, dx$$ 5. **Compute each integral:** $$\int f(x) \, dx = \int (x^3 - 6x^2 + 8x) \, dx = \frac{x^4}{4} - 2x^3 + 4x^2 + C$$ 6. **Evaluate the first integral:** $$\int_0^2 f(x) \, dx = \left[ \frac{x^4}{4} - 2x^3 + 4x^2 \right]_0^2 = \left( \frac{16}{4} - 16 + 16 \right) - 0 = (4 - 16 + 16) = 4$$ 7. **Evaluate the second integral:** $$\int_2^4 f(x) \, dx = \left[ \frac{x^4}{4} - 2x^3 + 4x^2 \right]_2^4 = \left( \frac{256}{4} - 128 + 64 \right) - \left( 4 - 16 + 16 \right) = (64 - 128 + 64) - 4 = 0 - 4 = -4$$ 8. **Calculate total area:** Since the second integral is negative, take its absolute value: $$\text{Area} = 4 + 4 = 8$$ **Final answer:** The total area of the shaded regions is $8$.