let f(x)=18log10(x) and g(x)=0.5x³ - 8x² + 34.5x - 27. let r and s be the two regions enclosed by the graphs…

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 $y = f(x)=18\log_{10}(x)$ and $y = g(x)=0.5x^{3}-8x^{2}+34.5x - 27$ are $x = 1$ and $x = 10$.

Step2: Set up the integral for the area

The sum of the areas of regions $R$ and $S$ is $A=\int_{1}^{10}|18\log_{10}(x)-(0.5x^{3}-8x^{2}+34.5x - 27)|dx$. Since we can use a graphing calculator, we can directly input the integral $\int_{1}^{10}|18\frac{\ln(x)}{\ln(10)}-(0.5x^{3}-8x^{2}+34.5x - 27)|dx$ into the calculator.

Answer:

Using a graphing calculator, the value of the integral $\int_{1}^{10}|18\frac{\ln(x)}{\ln(10)}-(0.5x^{3}-8x^{2}+34.5x - 27)|dx\approx34.667$