find the area of the region bounded by the graphs of the given equations.\ny = 5x - x^2, y = x\nthe area of…

Explanation:

Step1: Find intersection points

Set $5x - x^{2}=x$. Rearrange to $x^{2}-4x = 0$, factor out $x$: $x(x - 4)=0$. So $x = 0$ and $x = 4$ are the intersection - points.

Step2: Determine the upper and lower functions

For $0\leq x\leq4$, $y_1=5x - x^{2}$ is above $y_2=x$. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $f(x)=5x - x^{2}$, $g(x)=x$, $a = 0$, $b = 4$.

Step3: Calculate the integral

$A=\int_{0}^{4}[(5x - x^{2})-x]dx=\int_{0}^{4}(4x - x^{2})dx$. Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(4x - x^{2})dx=4\times\frac{x^{2}}{2}-\frac{x^{3}}{3}+C = 2x^{2}-\frac{x^{3}}{3}+C$. Evaluate the definite integral: $\left(2x^{2}-\frac{x^{3}}{3}\right)\big|_{0}^{4}=(2\times4^{2}-\frac{4^{3}}{3})-(2\times0^{2}-\frac{0^{3}}{3})=(32-\frac{64}{3})-0=\frac{96 - 64}{3}=\frac{32}{3}$.

Answer:

$\frac{32}{3}$