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

Explanation:

Step1: Determine the upper - function

On the interval $[-10,-4]$, we find which of $f(x)=x^{2}+8x + 16$ and $g(x)=8x + 80$ is the upper - function. Let $h(x)=g(x)-f(x)=(8x + 80)-(x^{2}+8x + 16)=64 - x^{2}$. When $x\in[-10,-4]$, $h(x)=64 - x^{2}\geq0$, so $g(x)$ is the upper - function and $f(x)$ is the lower - function.

Step2: Use the area formula

The area $A$ between two curves $y = g(x)$ and $y = f(x)$ over the interval $[a,b]$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here, $a=-10$, $b = - 4$, $g(x)=8x + 80$, and $f(x)=x^{2}+8x + 16$. Then $A=\int_{-10}^{-4}[(8x + 80)-(x^{2}+8x + 16)]dx=\int_{-10}^{-4}(64 - x^{2})dx$.

Step3: Integrate the function

We know that $\int(64 - x^{2})dx=64x-\frac{1}{3}x^{3}+C$.

Step4: Evaluate the definite integral

$A=\left[64x-\frac{1}{3}x^{3}\right]_{-10}^{-4}=\left(64\times(-4)-\frac{1}{3}\times(-4)^{3}\right)-\left(64\times(-10)-\frac{1}{3}\times(-10)^{3}\right)$. First, calculate $64\times(-4)-\frac{1}{3}\times(-4)^{3}=-256+\frac{64}{3}=\frac{-768 + 64}{3}=-\frac{704}{3}$. Second, calculate $64\times(-10)-\frac{1}{3}\times(-10)^{3}=-640+\frac{1000}{3}=\frac{-1920 + 1000}{3}=-\frac{920}{3}$. Then $A=-\frac{704}{3}-\left(-\frac{920}{3}\right)=\frac{-704 + 920}{3}=\frac{216}{3}=72$.

Answer:

72