find the area of the region r bounded by the graph of f and the x - axis on the given interval. graph f and…

Explanation:

Step1: Expand the function

$f(x)=x^{2}(x - 2)=x^{3}-2x^{2}$

Step2: Use the definite - integral formula for area

The area $A$ between the curve $y = f(x)$ and the $x$-axis on the interval $[a,b]$ is given by $A=\int_{a}^{b}|f(x)|dx$. First, find the zeros of $y = f(x)$: Set $f(x)=x^{3}-2x^{2}=x^{2}(x - 2)=0$, the zeros are $x = 0$ and $x = 2$. We need to split the integral based on the sign of $f(x)$ on $[-1,3]$. On $[-1,0]$, $f(x)=x^{3}-2x^{2}\leq0$, on $[0,2]$, $f(x)=x^{3}-2x^{2}\leq0$, on $[2,3]$, $f(x)=x^{3}-2x^{2}\geq0$. $A=-\int_{-1}^{0}(x^{3}-2x^{2})dx-\int_{0}^{2}(x^{3}-2x^{2})dx+\int_{2}^{3}(x^{3}-2x^{2})dx$

Step3: Integrate term - by - term

The antiderivative of $x^{3}-2x^{2}$ is $\frac{1}{4}x^{4}-\frac{2}{3}x^{3}+C$. $-\left[\frac{1}{4}x^{4}-\frac{2}{3}x^{3}\right]{-1}^{0}-\left[\frac{1}{4}x^{4}-\frac{2}{3}x^{3}\right]{0}^{2}+\left[\frac{1}{4}x^{4}-\frac{2}{3}x^{3}\right]{2}^{3}$ For $-\left[\frac{1}{4}x^{4}-\frac{2}{3}x^{3}\right]{-1}^{0}=-\left(0-\left(\frac{1}{4}+\frac{2}{3}\right)\right)=\frac{1}{4}+\frac{2}{3}=\frac{3 + 8}{12}=\frac{11}{12}$ For $-\left[\frac{1}{4}x^{4}-\frac{2}{3}x^{3}\right]{0}^{2}=-\left(\frac{1}{4}\times16-\frac{2}{3}\times8\right)=-\left(4-\frac{16}{3}\right)=\frac{4}{3}$ For $\left[\frac{1}{4}x^{4}-\frac{2}{3}x^{3}\right]{2}^{3}=\left(\frac{1}{4}\times81-\frac{2}{3}\times27\right)-\left(\frac{1}{4}\times16-\frac{2}{3}\times8\right)=\left(\frac{81}{4}-18\right)-\left(4 - \frac{16}{3}\right)=\frac{81}{4}-18 - 4+\frac{16}{3}=\frac{81}{4}-22+\frac{16}{3}=\frac{243 - 264+64}{12}=\frac{43}{12}$

Step4: Sum up the results

$A=\frac{11}{12}+\frac{4}{3}+\frac{43}{12}=\frac{11 + 16+43}{12}=\frac{70}{12}=\frac{35}{6}\approx5.83$

Answer:

$5.83$