Find the total area between \( f(x) \) and the \( x \)-axis from \( x=0 \) to \( x=6 \) \[ f(x)=x^{2}-x-12 \] Total Area \( =[ \) ?] Round your answer to the nearest thousandth.

Let \( f(x) = x^2 - x - 12 \). We want the total area between \( f(x) \) and the \( x \)-axis from \( x = 0 \) to \( x = 6 \). ### 1. Find the Roots Solve for \( x \) when \[ x^2 - x - 12 = 0. \] Factor: \[ (x - 4)(x + 3) = 0, \] so the roots are \[ x = 4 \quad \text{and} \quad x = -3. \] Within the interval \( [0, 6] \), the only relevant root is \( x = 4 \). ### 2. Determine the Sign of \( f(x) \) Test a point in each sub-interval: - For \( x \) in \( [0, 4] \), choose \( x = 2 \): \[ f(2) = 2^2 - 2 - 12 = 4 - 2 - 12 = -10 \quad (\text{negative}). \] - For \( x \) in \( [4, 6] \), choose \( x = 5 \): \[ f(5) = 5^2 - 5 - 12 = 25 - 5 - 12 = 8 \quad (\text{positive}). \] Thus, \( f(x) \) is below the \( x \)-axis from \( x = 0 \) to \( x = 4 \) and above the \( x \)-axis from \( x = 4 \) to \( x = 6 \). ### 3. Set Up the Integrals for the Total Area The total area is computed by integrating the absolute value: \[ \text{Total Area} = \int_{0}^{4} \lvert f(x) \rvert \, dx + \int_{4}^{6} \lvert f(x) \rvert \, dx. \] Since \( f(x) \) is negative on \( [0,4] \), \[ \lvert f(x) \rvert = -f(x) \quad \text{on } [0,4], \] and positive on \( [4,6] \), \[ \lvert f(x) \rvert = f(x) \quad \text{on } [4,6]. \] Thus, \[ \text{Total Area} = -\int_{0}^{4} (x^2 - x - 12) \, dx + \int_{4}^{6} (x^2 - x - 12) \, dx. \] ### 4. Compute the Antiderivative Find the antiderivative of \( f(x) \): \[ \int (x^2 - x - 12) \, dx = \frac{x^3}{3} - \frac{x^2}{2} - 12x + C. \] ### 5. Evaluate the Integrals #### First Integral: from \( x = 0 \) to \( x = 4 \) \[ -\int_{0}^{4} (x^2 - x - 12) \, dx = -\left[ \left( \frac{x^3}{3} - \frac{x^2}{2} - 12x \right) \Bigg|_{0}^{4} \right]. \] At \( x = 4 \): \[ \frac{4^3}{3} - \frac{4^2}{2} - 12(4) = \frac{64}{3} - \frac{16}{2} - 48 = \frac{64}{3} - 8 - 48 = \frac{64}{3} - 56. \] At \( x = 0 \), the antiderivative is \( 0 \). Thus, \[ -\left[\frac{64}{3} - 56\right] = 56 - \frac{64}{3} = \frac{168 - 64}{3} = \frac{104}{3}. \] #### Second Integral: from \( x = 4 \) to \( x = 6 \) \[ \int_{4}^{6} (x^2 - x - 12) \, dx = \left( \frac{x^3}{3} - \frac{x^2}{2} - 12x \right) \Bigg|_{4}^{6}. \] At \( x = 6 \): \[ \frac{6^3}{3} - \frac{6^2}{2} - 12(6) = \frac{216}{3} - \frac{36}{2} - 72 = 72 - 18 - 72 = -18. \] At \( x = 4 \) (computed previously): \[ \frac{4^3}{3} - \frac{4^2}{2} - 12(4) = \frac{64}{3} - 56. \] Thus, the second integral is: \[ -18 - \left(\frac{64}{3} - 56\right) = -18 - \frac{64}{3} + 56 = 38 - \frac{64}{3} = \frac{114 - 64}{3} = \frac{50}{3}. \] ### 6. Total Area Add the two areas: \[ \text{Total Area} = \frac{104}{3} + \frac{50}{3} = \frac{154}{3}. \] Rounded to the nearest thousandth: \[ \frac{154}{3} \approx 51.333. \] Thus, the total area between \( f(x) \) and the \( x \)-axis from \( x = 0 \) to \( x = 6 \) is approximately \( 51.333 \).