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 where $f(x)=x^{2}+8x + 16$ and $g(x)=x + 10$ intersect on the interval $[-10,-4]$. Set $f(x)=g(x)$: [x^{2}+8x + 16=x + 10] [x^{2}+7x+6 = 0] [(x + 1)(x+6)=0] The solutions are $x=-1$ and $x=-6$. On the interval $[-10,-6]$, $f(x)\geq g(x)$; on the interval $[-6,-4]$, $g(x)\geq f(x)$.

Step2: Set up the integral for the area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ and the $x$-axis on $[a,b]$ is given by: [A=\int_{-10}^{-6}(x^{2}+8x + 16)dx+\int_{-6}^{-4}(x + 10)dx]

Step3: Integrate the first integral

Integrate $\int(x^{2}+8x + 16)dx=\frac{1}{3}x^{3}+4x^{2}+16x+C$. [ \begin{align*} \int_{-10}^{-6}(x^{2}+8x + 16)dx&=\left[\frac{1}{3}x^{3}+4x^{2}+16x\right]_{-10}^{-6}\ &=\left(\frac{1}{3}(-6)^{3}+4(-6)^{2}+16(-6)\right)-\left(\frac{1}{3}(-10)^{3}+4(-10)^{2}+16(-10)\right)\ &=\left(-72 + 144-96\right)-\left(-\frac{1000}{3}+400 - 160\right)\ &=-24-\left(-\frac{1000}{3}+240\right)\ &=-24+\frac{1000}{3}-240\ &=\frac{-72 + 1000-720}{3}\ &=\frac{208}{3} \end{align*} ]

Step4: Integrate the second integral

Integrate $\int(x + 10)dx=\frac{1}{2}x^{2}+10x+C$. [ \begin{align*} \int_{-6}^{-4}(x + 10)dx&=\left[\frac{1}{2}x^{2}+10x\right]_{-6}^{-4}\ &=\left(\frac{1}{2}(-4)^{2}+10(-4)\right)-\left(\frac{1}{2}(-6)^{2}+10(-6)\right)\ &=(8-40)-(18 - 60)\ &=-32-(-42)\ &=10 \end{align*} ]

Step5: Calculate the total area

[A=\frac{208}{3}+10=\frac{208 + 30}{3}=\frac{238}{3}]

Answer:

$\frac{238}{3}$