Find the total area between the \( x \)-axis and the function on the given interval. \( f(x)=x^{3}-3 x^{2}+2 x \) \( [0,2] \) \( [?] \)

1. Factor the function: \[ f(x) = x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2) \] 2. Determine the zeros on \([0,2]\): The zeros are \( x = 0 \), \( x = 1 \), and \( x = 2 \). 3. Analyze the sign of \( f(x) \) on the intervals: - For \( x \in [0,1] \): - \( x > 0 \) - \( (x-1) < 0 \) - \( (x-2) < 0 \) Hence, \( f(x) = (+) \times (-) \times (-) > 0 \). - For \( x \in [1,2] \): - \( x > 0 \) - \( (x-1) > 0 \) - \( (x-2) < 0 \) Hence, \( f(x) = (+) \times (+) \times (-) < 0 \). 4. Set up the integral for total area: The total area \( A \) between the \( x \)-axis and \( f(x) \) is given by: \[ A = \int_{0}^{1} f(x) \, dx + \left| \int_{1}^{2} f(x) \, dx \right| \] 5. Compute the indefinite integral (antiderivative): \[ \int f(x) \, dx = \int \left( x^3 - 3x^2 + 2x \right) dx = \frac{x^4}{4} - x^3 + x^2 + C \] 6. Evaluate the integral from \( 0 \) to \( 1 \): \[ I_1 = \left[\frac{x^4}{4} - x^3 + x^2\right]_{0}^{1} = \left(\frac{1}{4} - 1 + 1\right) - \left(0\right) = \frac{1}{4} \] 7. Evaluate the integral from \( 1 \) to \( 2 \): \[ I_2 = \left[\frac{x^4}{4} - x^3 + x^2\right]_{1}^{2} \] For \( x = 2 \): \[ \frac{2^4}{4} - 2^3 + 2^2 = \frac{16}{4} - 8 + 4 = 4 - 8 + 4 = 0 \] For \( x = 1 \): \[ \frac{1}{4} - 1 + 1 = \frac{1}{4} \] Thus, \[ I_2 = 0 - \frac{1}{4} = -\frac{1}{4} \] 8. Compute the total area: \[ A = I_1 + \left| I_2 \right| = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \] Therefore, the total area between the \( x \)-axis and the function on \([0,2]\) is \( \frac{1}{2} \).