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

Explanation:

Step1: Determine which function is greater

Find the intersection of $f(x)=x^{2}+4x + 4$ and $g(x)=-x + 4$ by setting $x^{2}+4x + 4=-x + 4$. $x^{2}+4x + 4+x - 4=0$, $x^{2}+5x=0$, $x(x + 5)=0$. So $x = 0$ or $x=-5$. In the interval $[-2,4]$, we test a value, say $x = 0$. $f(0)=4$ and $g(0)=4$. For $x\in[-2,0]$, $f(x)\geq g(x)$; for $x\in[0,4]$, $g(x)\geq f(x)$.

Step2: Set up the integral for the area

The area $A=\int_{-2}^{0}(x^{2}+4x + 4)dx+\int_{0}^{4}(-x + 4)dx$.

Step3: Integrate the first - integral

$\int_{-2}^{0}(x^{2}+4x + 4)dx=\left[\frac{1}{3}x^{3}+2x^{2}+4x\right]_{-2}^0$. $0-\left(\frac{1}{3}(-2)^{3}+2(-2)^{2}+4(-2)\right)=0-\left(-\frac{8}{3}+8 - 8\right)=\frac{8}{3}$.

Step4: Integrate the second - integral

$\int_{0}^{4}(-x + 4)dx=\left[-\frac{1}{2}x^{2}+4x\right]_{0}^{4}$. $\left(-\frac{1}{2}(4)^{2}+4\times4\right)-0=-8 + 16=8$.

Step5: Calculate the total area

$A=\frac{8}{3}+8=\frac{8 + 24}{3}=\frac{32}{3}$.

Answer:

$\frac{32}{3}$