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

To find the total area between the \( x \)-axis and the function \( f(x) = x^{3}-3x^{2}+2x \) on the interval \([0,2]\), we first need to determine where the function crosses the \( x \)-axis by finding the roots of the equation \( f(x) = 0 \). Factoring, we see that \( f(x) = x(x-1)(x-2) \). The roots are \( x = 0, 1, \) and \( 2 \). Next, we need to calculate the area from \( x = 0 \) to \( x = 2 \). We'll split the interval into separate parts because the function is below the \( x \)-axis between \( x = 1 \) and \( x = 2 \): 1. From \( x = 0 \) to \( x = 1 \): The area is \( \int_0^1 (x^{3}-3x^{2}+2x) \,dx \). 2. From \( x = 1 \) to \( x = 2 \): The area is \( \int_1^2 -(x^{3}-3x^{2}+2x) \,dx \) (we take the negative since the function is below the axis). Calculating these integrals: 1. For \( \int_0^1 (x^{3}-3x^{2}+2x) \,dx \): \[ = \left[ \frac{x^{4}}{4}-x^{3}+\frac{2x^{2}}{2} \right]_0^1 = \left[ \frac{1}{4} - 1 + 1 \right] - [0] = \frac{1}{4}. \] 2. For \( \int_1^2 -(x^{3}-3x^{2}+2x) \,dx \): \[ = -\left[ \frac{x^{4}}{4}-x^{3}+\frac{2x^{2}}{2} \right]_1^2 = -\left[ \left( \frac{16}{4} - 8 + 8 \right) - \left( \frac{1}{4} - 1 + 1 \right) \right] = -\left[ 4 - \frac{1}{4} \right] = -\left[ \frac{16}{4} - \frac{1}{4} \right] = -\frac{15}{4}. \] The total area is then: \[ \text{Total Area} = \frac{1}{4} + \frac{15}{4} = \frac{16}{4} = 4. \] Thus, the total area between the \( x \)-axis and the function \( f(x) \) on the interval \([0, 2]\) is \( 4 \).