what is the area of the region between the graphs of f(x) = 4/x and g(x) = 5 from x = -6 to x = -2? choose 1…

Explanation:

Step1: Set up the integral

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)=\frac{4}{x}$, $g(x) = 5$, $a=-6$ and $b = - 2$. Since for $x\in[-6,-2]$, $5>\frac{4}{x}$, the integral is $A=\int_{-6}^{-2}(5 - \frac{4}{x})dx$.

Step2: Integrate term - by - term

We know that $\int(5-\frac{4}{x})dx=\int 5dx-\int\frac{4}{x}dx$. Using the power rule $\int kdx=kx + C$ ($k$ is a constant) and $\int\frac{1}{x}dx=\ln|x|+C$, we have $\int 5dx=5x$ and $\int\frac{4}{x}dx = 4\ln|x|$. So, $\int_{-6}^{-2}(5 - \frac{4}{x})dx=\left[5x-4\ln|x|\right]_{-6}^{-2}$.

Step3: Evaluate the definite integral

[ \begin{align*} \left[5x-4\ln|x|\right]_{-6}^{-2}&=(5\times(-2)-4\ln|-2|)-(5\times(-6)-4\ln|-6|)\ &=(-10 - 4\ln2)-(-30-4\ln6)\ &=-10 - 4\ln2 + 30+4\ln6\ &=20+4(\ln6-\ln2)\ &=20 + 4\ln\frac{6}{2}\ &=20+4\ln3\ &=20+\ln(3^4)\ &=20+\ln(81) \end{align*} ]

Answer:

C. $20+\ln(81)$