let $r$ be the region bounded by the graphs of $y = 2x$ and $y = 4x - x^{2}$. what is the area of $r$?\n(a)…

Explanation:

Step1: Find intersection points

Set $2x = 4x - x^{2}$, then $x^{2}-2x = 0$, factoring gives $x(x - 2)=0$. 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, $f(x)=4x - x^{2}$, $g(x)=2x$, $a = 0$, $b = 2$.

Step3: Calculate the integral

$A=\int_{0}^{2}[(4x - x^{2})-2x]dx=\int_{0}^{2}(2x - x^{2})dx$. Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int_{0}^{2}(2x - x^{2})dx=\left[x^{2}-\frac{1}{3}x^{3}\right]_{0}^{2}$.

Step4: Evaluate the definite integral

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

Answer:

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