1. **State the problem:** We need to find the total area of the shaded regions enclosed by the curve $$y = 12x^3 - 48x^2 + 36x$$ and the x-axis, between the roots $$x=0$$, $$x=1$$, and $$x=3$$. 2. **Formula and rules:** The total area between a curve and the x-axis over an interval is given by the integral of the absolute value of the function: $$\text{Area} = \int_a^b |f(x)| \, dx$$ Since the curve crosses the x-axis at $$x=0$$, $$x=1$$, and $$x=3$$, we split the integral at these points and take the absolute value of each integral over the intervals where the function is positive or negative. 3. **Determine the sign of the function on each interval:** - For $$0 < x < 1$$, the curve is above the x-axis, so $$f(x) > 0$$. - For $$1 < x < 3$$, the curve is below the x-axis, so $$f(x) < 0$$. 4. **Calculate the definite integrals:** Calculate $$\int_0^1 (12x^3 - 48x^2 + 36x) \, dx$$: $$\int 12x^3 \, dx = 3x^4$$ $$\int -48x^2 \, dx = -16x^3$$ $$\int 36x \, dx = 18x^2$$ So, $$\int_0^1 (12x^3 - 48x^2 + 36x) \, dx = \left[3x^4 - 16x^3 + 18x^2\right]_0^1 = (3 - 16 + 18) - 0 = 5$$ Calculate $$\int_1^3 (12x^3 - 48x^2 + 36x) \, dx$$: $$\left[3x^4 - 16x^3 + 18x^2\right]_1^3 = (3(3)^4 - 16(3)^3 + 18(3)^2) - (3(1)^4 - 16(1)^3 + 18(1)^2)$$ Calculate each term: $$3(3)^4 = 3 \times 81 = 243$$ $$-16(3)^3 = -16 \times 27 = -432$$ $$18(3)^2 = 18 \times 9 = 162$$ Sum for $$x=3$$: $$243 - 432 + 162 = -27$$ Sum for $$x=1$$: $$3 - 16 + 18 = 5$$ So, $$-27 - 5 = -32$$ 5. **Calculate total area:** Since the function is negative between 1 and 3, take the absolute value of the integral over that interval. Total area = $$5 + | -32 | = 5 + 32 = 37$$ **Final answer:** $$\boxed{37}$$