current objective find the area of a region bounded by two functions that cross question determine the area…

Explanation:

Step1: Find the difference of functions

Let $h(x)=f(x)-g(x)=(x^{3}+19x^{2}+11x)-(6x^{2}-19x)=x^{3}+13x^{2}+30x$.

Step2: Use the definite - integral formula for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ on the interval $[a,b]$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $a=-2$, $b = 2$. So $A=\int_{-2}^{2}|x^{3}+13x^{2}+30x|dx$. Since $y=x^{3}+30x$ is an odd - function ($y(-x)=(-x)^{3}+30(-x)=-(x^{3}+30x)$) and $y = 13x^{2}$ is an even - function ($y(-x)=13(-x)^{2}=13x^{2}$), we know that $\int_{-2}^{2}(x^{3}+30x)dx = 0$ and $\int_{-2}^{2}13x^{2}dx$.

Step3: Calculate the definite integral of the even part

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int_{-2}^{2}13x^{2}dx=2\int_{0}^{2}13x^{2}dx$. First, $\int 13x^{2}dx=\frac{13}{3}x^{3}+C$. Then $2\int_{0}^{2}13x^{2}dx=2\times[\frac{13}{3}x^{3}]_{0}^{2}=2\times\frac{13}{3}\times(2^{3}-0^{3})=\frac{208}{3}$.

Answer:

$\frac{208}{3}$