Question

question. determine the area, in square units, of the region bounded by f(x)=x^3 + 19x^2+11x and g(x)=6x^2 - 19x over the interval -2,2. provide your answer below.

Explanation:

Step1: Find the difference of the 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 given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here $a=-2$, $b = 2$ and $A=\int_{-2}^{2}(x^{3}+13x^{2}+30x)dx$.
Since $\int_{-2}^{2}(x^{3}+30x)dx = 0$ (because $y=x^{3}+30x$ is an odd function and $\int_{-a}^{a}f(x)dx = 0$ for an odd function $f(x)$), then $A=\int_{-2}^{2}13x^{2}dx$.

Step3: Evaluate the definite integral

We know that $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). So $\int_{-2}^{2}13x^{2}dx=13\times\frac{x^{3}}{3}\big|_{-2}^{2}=13\times(\frac{2^{3}}{3}-\frac{(-2)^{3}}{3})$.
$=13\times\frac{8 + 8}{3}=13\times\frac{16}{3}=\frac{208}{3}$.

Answer:

$\frac{208}{3}$