current objective find the area of a region bounded above by two different functions question determine the…

Explanation:

Step1: Determine the upper - function

We need to find which of $f(x)=-x^{2}-8x$ and $g(x)=3x + 24$ is greater on the interval $[-8,0]$. Let's find the intersection of $f(x)$ and $g(x)$: Set $-x^{2}-8x=3x + 24$. Rearrange to get $x^{2}+11x + 24=0$. Factor: $(x + 3)(x+8)=0$. The solutions are $x=-3$ and $x=-8$. We can test a value in the sub - intervals. For $x=-4$: $f(-4)=-(-4)^{2}-8\times(-4)=-16 + 32 = 16$ and $g(-4)=3\times(-4)+24=-12 + 24 = 12$. So, on $[-8,-3]$, $g(x)\geq f(x)$ and on $[-3,0]$, $f(x)\geq g(x)$.

Step2: Set up the integral for the area

The area $A$ is given by: $A=\int_{-8}^{-3}[(3x + 24)-0]dx+\int_{-3}^{0}[(-x^{2}-8x)-0]dx$.

Step3: Integrate the first integral

$\int_{-8}^{-3}(3x + 24)dx=\left[\frac{3x^{2}}{2}+24x\right]_{-8}^{-3}$. $=\left(\frac{3(-3)^{2}}{2}+24\times(-3)\right)-\left(\frac{3(-8)^{2}}{2}+24\times(-8)\right)$. $=\left(\frac{27}{2}-72\right)-\left(96 - 192\right)$. $=\frac{27}{2}-72 - 96 + 192$. $=\frac{27}{2}+24=\frac{27 + 48}{2}=\frac{75}{2}$.

Step4: Integrate the second integral

$\int_{-3}^{0}(-x^{2}-8x)dx=\left[-\frac{x^{3}}{3}-4x^{2}\right]_{-3}^{0}$. $=0-\left(-\frac{(-3)^{3}}{3}-4\times(-3)^{2}\right)$. $=-\left(9 - 36\right)=27$.

Step5: Calculate the total area

$A=\frac{75}{2}+27=\frac{75+54}{2}=\frac{129}{2}$.

Answer:

$\frac{129}{2}$