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

Explanation:

Step1: Determine the upper - function

First, we need to find which of $f(x)=x^{2}-2x + 1$ and $g(x)=8x + 40$ is the upper - function on the interval $[-5,1]$. We can find the difference $h(x)=g(x)-f(x)=(8x + 40)-(x^{2}-2x + 1)=-x^{2}+10x + 39$. We can also check the values of $f(x)$ and $g(x)$ at the endpoints of the interval. $f(-5)=(-5)^{2}-2\times(-5)+1=25 + 10+1=36$, $g(-5)=8\times(-5)+40=0$. $f(1)=1^{2}-2\times1 + 1=0$, $g(1)=8\times1+40=48$. So, $g(x)$ is the upper - function on $[-5,1]$.

Step2: Use the definite - integral formula for area

The area $A$ between a function $y = g(x)$ and the $x$ - axis over the interval $[a,b]$ is given by $A=\int_{a}^{b}g(x)dx$ when $g(x)\geq0$ on $[a,b]$. Here, $a=-5$, $b = 1$, and $g(x)=8x + 40$. We know that $\int(8x + 40)dx=8\times\frac{x^{2}}{2}+40x=4x^{2}+40x+C$.

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus $\int_{a}^{b}g(x)dx=\left[4x^{2}+40x\right]_{-5}^{1}$. First, substitute $x = 1$: $4\times1^{2}+40\times1=4 + 40=44$. Then substitute $x=-5$: $4\times(-5)^{2}+40\times(-5)=4\times25-200=100 - 200=-100$. Then $A=44-(-100)=44 + 100=144$.

Answer:

144