let f(x)=3x³ - 8x² - 3x + 12 and g(x)=4e^(x + 1)(x - 3)+2x + 2. let r and s be the two regions enclosed by…

let f(x)=3x³ - 8x² - 3x + 12 and g(x)=4e^(x + 1)(x - 3)+2x + 2. let r and s be the two regions enclosed by the graphs of f and g as shown in the graph. find the sum of the areas of regions r and s. use a graphing calculator and round your answer to three decimal places.

let f(x)=3x³ - 8x² - 3x + 12 and g(x)=4e^(x + 1)(x - 3)+2x + 2. let r and s be the two regions enclosed by the graphs of f and g as shown in the graph. find the sum of the areas of regions r and s. use a graphing calculator and round your answer to three decimal places.

Answer

Explanation:

Step1: Recall area - between - curves formula

The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. The intersection points of the curves $f(x)=3x^{3}-8x^{2}-3x + 12$ and $g(x)=4e^{(x + 1)(x - 3)}+2x + 2$ are $x=-1$ and $x = 3$.

Step2: Set up the integral for the area

The sum of the areas of regions $R$ and $S$ is $A=\int_{-1}^{3}|(3x^{3}-8x^{2}-3x + 12)-(4e^{(x + 1)(x - 3)}+2x + 2)|dx=\int_{-1}^{3}|3x^{3}-8x^{2}-5x + 10-4e^{x^{2}-2x - 3}|dx$.

Step3: Use a graphing - calculator

Using a graphing calculator (such as TI - 84 Plus: enter $Y_1=3x^{3}-8x^{2}-5x + 10-4e^{x^{2}-2x - 3}$, and then use the integral function $\int_{-1}^{3}Y_1dx$).

Answer:

$31.637$