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

Explanation:

Step1: Determine the upper - function

We need to find which of $f(x)=x^{2}-6x + 9$ and $g(x)=-3x + 27$ is greater on the interval $[3,9]$. Let $h(x)=f(x)-g(x)=x^{2}-6x + 9-(-3x + 27)=x^{2}-3x - 18$. Factor $h(x)$: $h(x)=(x - 6)(x+3)$. The roots of $h(x)$ are $x=-3$ and $x = 6$. On the interval $[3,6)$, $h(x)<0$, so $g(x)>f(x)$; on the interval $(6,9]$, $h(x)>0$, so $f(x)>g(x)$.

Step2: Set up the integral for the area

The area $A$ of the region $R$ is given by $A=\int_{3}^{6}g(x)dx+\int_{6}^{9}f(x)dx$. First, calculate $\int_{3}^{6}(-3x + 27)dx$. Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-3x + 27)dx=-\frac{3}{2}x^{2}+27x+C$. Then $\int_{3}^{6}(-3x + 27)dx=\left(-\frac{3}{2}(6)^{2}+27\times6\right)-\left(-\frac{3}{2}(3)^{2}+27\times3\right)=\left(-54 + 162\right)-\left(-\frac{27}{2}+81\right)=108-(81-\frac{27}{2})=108 - 81+\frac{27}{2}=27+\frac{27}{2}=\frac{54 + 27}{2}=\frac{81}{2}$. Second, calculate $\int_{6}^{9}(x^{2}-6x + 9)dx$. Using the power - rule, $\int(x^{2}-6x + 9)dx=\frac{1}{3}x^{3}-3x^{2}+9x+C$. Then $\int_{6}^{9}(x^{2}-6x + 9)dx=\left(\frac{1}{3}(9)^{3}-3(9)^{2}+9\times9\right)-\left(\frac{1}{3}(6)^{3}-3(6)^{2}+9\times6\right)=\left(243-243 + 81\right)-\left(72-108 + 54\right)=81-(18)=63$.

Step3: Calculate the total area

$A=\frac{81}{2}+63=\frac{81+126}{2}=\frac{207}{2}$.

Answer:

$\frac{207}{2}$