sketch the region enclosed by the given curves. (y = 3/x), (y = 12x), (y=\frac{1}{3}x), (x > 0). find its…

Explanation:

Step1: Find intersection points

Set $\frac{3}{x}=12x$, then $12x^{2}=3$, $x^{2}=\frac{1}{4}$, since $x > 0$, $x=\frac{1}{2}$. Set $\frac{3}{x}=\frac{1}{3}x$, then $x^{2}=9$, since $x>0$, $x = 3$.

Step2: Set up integral for area

The area $A=\int_{0}^{\frac{1}{2}}(12x-\frac{1}{3}x)dx+\int_{\frac{1}{2}}^{3}(\frac{3}{x}-\frac{1}{3}x)dx$.

Step3: Integrate first - part

$\int_{0}^{\frac{1}{2}}(12x-\frac{1}{3}x)dx=\int_{0}^{\frac{1}{2}}\frac{36x - x}{3}dx=\int_{0}^{\frac{1}{2}}\frac{35x}{3}dx=\left[\frac{35x^{2}}{6}\right]_{0}^{\frac{1}{2}}=\frac{35}{24}$.

Step4: Integrate second - part

$\int_{\frac{1}{2}}^{3}(\frac{3}{x}-\frac{1}{3}x)dx=\left[3\ln x-\frac{x^{2}}{6}\right]_{\frac{1}{2}}^{3}=3\ln3-\frac{3}{2}-3\ln\frac{1}{2}+\frac{1}{24}=3\ln6-\frac{35}{24}$.

Step5: Calculate total area

$A=\frac{35}{24}+3\ln6-\frac{35}{24}=3\ln6$.

Answer:

$3\ln6$