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

Explanation:

Step1: Find the difference of the functions

First, find $f(x)-g(x)$: [ \begin{align*} f(x)-g(x)&=(8x^{3}+7x^{2}+28x + 6)-(7x^{3}-6x^{2}-8x + 6)\ &=8x^{3}+7x^{2}+28x + 6 - 7x^{3}+6x^{2}+8x - 6\ &=(8x^{3}-7x^{3})+(7x^{2}+6x^{2})+(28x + 8x)+(6 - 6)\ &=x^{3}+13x^{2}+36x \end{align*} ]

Step2: Use the definite - integral formula for the area between two curves

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ over the interval $[a,b]$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Since we are integrating $h(x)=x^{3}+13x^{2}+36x$ over $[-2,2]$, we calculate $\int_{-2}^{2}(x^{3}+13x^{2}+36x)dx$. We know that $\int_{-2}^{2}(x^{3}+36x)dx = 0$ because $y = x^{3}$ and $y = 36x$ are odd functions and for an odd function $y = h(x)$ and the interval $[-a,a]$, $\int_{-a}^{a}h(x)dx=0$. So, $\int_{-2}^{2}(x^{3}+13x^{2}+36x)dx=\int_{-2}^{2}13x^{2}dx$.

Step3: Integrate the remaining function

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int 13x^{2}dx=13\times\frac{x^{3}}{3}+C$. Then, $\int_{-2}^{2}13x^{2}dx=13\times\frac{x^{3}}{3}\big|_{-2}^{2}=13\times(\frac{2^{3}}{3}-\frac{(-2)^{3}}{3})$. [ \begin{align*} &13\times(\frac{8}{3}+\frac{8}{3})\ &=13\times\frac{16}{3}\ &=\frac{208}{3} \end{align*} ]

Answer:

$\frac{208}{3}$