9 mark for review let r be the region bounded by the graphs of y = 2x and y = 4x - x². what is the area of…

Explanation:

Step1: Find intersection points

Set $2x = 4x - x^{2}$. Rearrange to $x^{2}-2x = 0$, factor out $x$: $x(x - 2)=0$. So $x = 0$ and $x = 2$ are the intersection - points.

Step2: Determine the upper and lower functions

For $0\leq x\leq2$, $y_1=4x - x^{2}$ is above $y_2 = 2x$. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}(f(x)-g(x))dx$. Here, $A=\int_{0}^{2}((4x - x^{2})-2x)dx=\int_{0}^{2}(2x - x^{2})dx$.

Step3: Integrate

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

Step4: Evaluate the definite - integral

$\left(2^{2}-\frac{1}{3}\times2^{3}\right)-\left(0^{2}-\frac{1}{3}\times0^{3}\right)=4-\frac{8}{3}=\frac{12 - 8}{3}=\frac{4}{3}$.

Answer:

B. $\frac{4}{3}$