find the total area between f(x) and the x - axis from x = 0 to x = 6.\nf(x)=x^{2}-x - 12\ntotal area =…

Explanation:

Step1: Find the roots of the function

Set $f(x)=x^{2}-x - 12 = 0$. Factoring gives $(x - 4)(x+3)=0$, so the roots are $x = 4$ and $x=-3$. We are interested in the interval $[0,6]$.

Step2: Determine where the function is above or below the x - axis

We can test points in the sub - intervals of $[0,6]$. Let's test $x = 1$ in $f(x)$: $f(1)=1^{2}-1 - 12=-12<0$. Let's test $x = 5$: $f(5)=5^{2}-5 - 12=8>0$. So $f(x)$ is negative on $[0,4)$ and positive on $(4,6]$.

Step3: Calculate the definite integrals

The area $A$ is given by $A=\left|\int_{0}^{4}(x^{2}-x - 12)dx\right|+\int_{4}^{6}(x^{2}-x - 12)dx$. First, find $\int(x^{2}-x - 12)dx=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-12x + C$. For $\int_{0}^{4}(x^{2}-x - 12)dx=\left[\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-12x\right]{0}^{4}=\frac{1}{3}(4)^{3}-\frac{1}{2}(4)^{2}-12(4)=\frac{64}{3}-8 - 48=\frac{64}{3}-56=\frac{64 - 168}{3}=-\frac{104}{3}$. $\left|\int{0}^{4}(x^{2}-x - 12)dx\right|=\frac{104}{3}$. For $\int_{4}^{6}(x^{2}-x - 12)dx=\left[\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-12x\right]_{4}^{6}=(\frac{1}{3}(6)^{3}-\frac{1}{2}(6)^{2}-12(6))-(\frac{1}{3}(4)^{3}-\frac{1}{2}(4)^{2}-12(4))=(72 - 18-72)-(\frac{64}{3}-8 - 48)=-18-(\frac{64 - 24 - 144}{3})=-18+\frac{104}{3}=\frac{-54 + 104}{3}=\frac{50}{3}$.

Step4: Sum the areas

$A=\frac{104}{3}+\frac{50}{3}=\frac{154}{3}\approx51.333$.

Answer:

$51.333$