let r be the region bounded by the functions f(x)=2x² + 4x and g(x)= - 4x² - 10x as shown in the diagram…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $2x^{2}+4x=-4x^{2}-10x$. Combine like - terms: $6x^{2}+14x = 0$. Factor out $2x$: $2x(3x + 7)=0$. Solve for $x$: $x = 0$ or $x=-\frac{7}{3}$.

Step2: Determine the upper - lower functions

On the interval $[-\frac{7}{3},0]$, $f(x)\geq g(x)$. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a =-\frac{7}{3}$, $b = 0$, and $f(x)-g(x)=(2x^{2}+4x)-(-4x^{2}-10x)=6x^{2}+14x$.

Step3: Calculate the integral

$A=\int_{-\frac{7}{3}}^{0}(6x^{2}+14x)dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(6x^{2}+14x)dx=6\times\frac{x^{3}}{3}+14\times\frac{x^{2}}{2}=2x^{3}+7x^{2}$. Evaluate the definite integral: $A=\left[2x^{3}+7x^{2}\right]_{-\frac{7}{3}}^{0}$. Substitute the upper and lower limits: $A = 0-(2\times(-\frac{7}{3})^{3}+7\times(-\frac{7}{3})^{2})$. $2\times(-\frac{7}{3})^{3}=2\times(-\frac{343}{27})=-\frac{686}{27}$, $7\times(-\frac{7}{3})^{2}=7\times\frac{49}{9}=\frac{343}{9}=\frac{1029}{27}$. Then $A=-\left(-\frac{686}{27}+\frac{1029}{27}\right)=\frac{343}{27}\approx12.704$.

Answer:

$12.704$