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.

We first find where the function \[ f(x)=x^2-x-12 \] crosses the \( x \)-axis by solving \[ x^2 - x - 12 = 0. \] ### Step 1. Find the Roots Factor the quadratic: \[ x^2 - x - 12 = (x-4)(x+3)=0. \] Thus, the roots are: \[ x=4 \quad \text{and} \quad x=-3. \] Since we are interested in the interval from \( x=0 \) to \( x=6 \), only \( x=4 \) is relevant. ### Step 2. Determine the Sign of \( f(x) \) Evaluate the function at a point between \( x=0 \) and \( x=4 \), for example, \( x=0 \): \[ f(0)=0^2-0-12=-12. \] Since \( f(0) < 0 \), the function is below the \( x \)-axis on the interval from \( x=0 \) to \( x=4 \). Next, evaluate at a point between \( x=4 \) and \( x=6 \), for example, \( x=6 \): \[ f(6)=6^2-6-12=36-6-12=18. \] This tells us that \( f(x) \) is above the \( x \)-axis from \( x=4 \) to \( x=6 \). ### Step 3. Set Up the Integrals for Total Area The total area between the curve and the \( x \)-axis is the sum of the area below the axis from \( x=0 \) to \( x=4 \) (taking the absolute value) and the area above the axis from \( x=4 \) to \( x=6 \). The antiderivative of \( f(x) \) is: \[ F(x)=\int (x^2-x-12)\,dx=\frac{x^3}{3}-\frac{x^2}{2}-12x. \] #### Area from \( x=0 \) to \( x=4 \): Since \( f(x) \) is negative on this interval, \[ \text{Area}_1 = -\int_{0}^{4}(x^2-x-12)\,dx. \] Compute \( F(4) \): \[ F(4)=\frac{4^3}{3}-\frac{4^2}{2}-12(4)=\frac{64}{3}-\frac{16}{2}-48=\frac{64}{3}-8-48=\frac{64}{3}-56. \] Compute \( F(0) \): \[ F(0)=0. \] Thus, \[ \int_{0}^{4}(x^2-x-12)\,dx = F(4)-F(0)=\frac{64}{3}-56. \] Since this value is negative, the area is: \[ \text{Area}_1 = -\left(\frac{64}{3}-56\right)=56-\frac{64}{3}=\frac{168}{3}-\frac{64}{3}=\frac{104}{3}. \] #### Area from \( x=4 \) to \( x=6 \): On this interval, \( f(x) \) is positive, so the area is: \[ \text{Area}_2 = \int_{4}^{6}(x^2-x-12)\,dx. \] Compute \( F(6) \): \[ F(6)=\frac{6^3}{3}-\frac{6^2}{2}-12(6)=\frac{216}{3}-\frac{36}{2}-72=72-18-72=-18. \] Using the previously computed \( F(4) \): \[ F(4)=\frac{64}{3}-56. \] Thus, \[ \int_{4}^{6}(x^2-x-12)\,dx = F(6)-F(4)=-18-\left(\frac{64}{3}-56\right)=-18-\frac{64}{3}+56. \] Simplify: \[ -18 + 56 = 38, \] so, \[ \text{Area}_2 = 38-\frac{64}{3}=\frac{114}{3}-\frac{64}{3}=\frac{50}{3}. \] ### Step 4. Compute the Total Area The total area is: \[ \text{Total Area} = \text{Area}_1+\text{Area}_2 = \frac{104}{3}+\frac{50}{3}=\frac{154}{3}. \] Convert to a decimal: \[ \frac{154}{3} \approx 51.333. \] ### Final Answer \[ \boxed{51.333} \]