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

Explanation:

Step1: Determine the upper - function

First, find which function is greater on the interval $[-5,-1]$. Let $h(x)=f(x)-g(x)=x^{2}+2x + 1-(2x + 10)=x^{2}-9=(x + 3)(x - 3)$. On the interval $[-5,-3]$, $f(x)\geq g(x)$; on the interval $[-3,-1]$, $g(x)\geq f(x)$.

Step2: Set up the integral for the area

The area $A$ is given by the sum of two definite - integrals: [A=\int_{-5}^{-3}(x^{2}+2x + 1)dx+\int_{-3}^{-1}(2x + 10)dx]

Step3: Integrate the first integral

Integrate $\int_{-5}^{-3}(x^{2}+2x + 1)dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: [\int(x^{2}+2x + 1)dx=\frac{x^{3}}{3}+x^{2}+x+C] [ \left[\frac{x^{3}}{3}+x^{2}+x\right]_{-5}^{-3}=\left(\frac{(-3)^{3}}{3}+(-3)^{2}+(-3)\right)-\left(\frac{(-5)^{3}}{3}+(-5)^{2}+(-5)\right)] [=\left(-9 + 9-3\right)-\left(-\frac{125}{3}+25 - 5\right)] [=-3+\frac{125}{3}-20] [=\frac{-9 + 125-60}{3}=\frac{56}{3}]

Step4: Integrate the second integral

Integrate $\int_{-3}^{-1}(2x + 10)dx$. Using the power rule, $\int(2x + 10)dx=x^{2}+10x+C$. [ \left[x^{2}+10x\right]_{-3}^{-1}=\left((-1)^{2}+10\times(-1)\right)-\left((-3)^{2}+10\times(-3)\right)] [=(1 - 10)-(9-30)] [=-9 + 21=12]

Step5: Calculate the total area

[A=\frac{56}{3}+12=\frac{56 + 36}{3}=\frac{92}{3}]

Answer:

$\frac{92}{3}$