let r be the region bounded by the functions f(x)=x² and g(x)= -x² - 2x as shown in the diagram below. find…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $x^{2}=-x^{2}-2x$. Rearranging gives $2x^{2}+2x = 0$, factoring out $2x$ we get $2x(x + 1)=0$. The solutions are $x=-1$ and $x = 0$.

Step2: Determine the upper - lower functions

On the interval $[-1,0]$, $f(x)\geq g(x)$. 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, $a=-1$, $b = 0$, $f(x)=x^{2}$ and $g(x)=-x^{2}-2x$. So $A=\int_{-1}^{0}[x^{2}-(-x^{2}-2x)]dx=\int_{-1}^{0}(2x^{2}+2x)dx$.

Step3: Integrate the function

Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(2x^{2}+2x)dx=2\times\frac{x^{3}}{3}+2\times\frac{x^{2}}{2}=\frac{2}{3}x^{3}+x^{2}+C$.

Step4: Evaluate the definite integral

$A=\left[\frac{2}{3}x^{3}+x^{2}\right]_{-1}^{0}$. Substitute the upper and lower limits: $A=(0)-\left(\frac{2}{3}\times(-1)^{3}+(-1)^{2}\right)=0-\left(-\frac{2}{3}+1\right)=0-\frac{1}{3}=-\frac{1}{3}$. Since area is non - negative, $A=\frac{1}{3}$.

Answer:

$\frac{1}{3}$