let r be the region bounded by the functions f(x)=3x² + 3x and g(x)= - 3x² + 9x as shown in the diagram…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $3x^{2}+3x=-3x^{2}+9x$. Rearrange to get $6x^{2}-6x = 0$, factor out $6x$: $6x(x - 1)=0$. The solutions are $x = 0$ and $x = 1$.

Step2: Determine the upper - lower function

For $x\in[0,1]$, we check $f(x)-g(x)=(3x^{2}+3x)-(-3x^{2}+9x)=6x^{2}-6x$. When $x\in[0,1]$, $g(x)\geq f(x)$. The area $A$ between two curves $y = g(x)$ and $y = f(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here $a = 0$, $b = 1$, and $g(x)-f(x)=(-3x^{2}+9x)-(3x^{2}+3x)=-6x^{2}+6x$.

Step3: Calculate the definite integral

$A=\int_{0}^{1}(-6x^{2}+6x)dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: $\int(-6x^{2}+6x)dx=-6\times\frac{x^{3}}{3}+6\times\frac{x^{2}}{2}=-2x^{3}+3x^{2}+C$. Evaluating the definite integral: $A=\left[-2x^{3}+3x^{2}\right]_{0}^{1}$. $A=(-2\times1^{3}+3\times1^{2})-(-2\times0^{3}+3\times0^{2})$. $A=-2 + 3=1$.

Answer:

$1$