question_answer

The area of region \[(x,y):\,\]\[xy\,\le \,8,\]\[\,1\,\le \,y\,\le \,{{x}^{2}}\] is:

A) \[16\,\,\text{lo}{{\text{g}}_{e}}\,2\,\,-\,\frac{14}{3}\]     

B) \[\text{lo}{{\text{g}}_{e}}\,2\,\,-\,\frac{7}{3}\]

C) \[8\,\,\text{lo}{{\text{g}}_{e}}\,2\,-\,\frac{14}{3}\]                     

D) \[8\,\,\text{lo}{{\text{g}}_{e}}\,2\,-\,6\]  

Correct Answer: A

Solution :

\[xy\ge 8\,\text{and}\,1\le y\,\le {{x}^{2}}\]

\[\text{A}=\int\limits_{1}^{2}{({{x}^{2}}-1)dx+\int\limits_{2}^{8}{\left( \frac{8}{x}\,-\,1 \right)\text{d}x}}\]

\[{{\left. \text{A}=\frac{{{x}^{3}}}{3} \right|}^{2}}_{1}+8\,\text{In}\left. x \right|_{2}^{8}-1-6\]

\[\text{A}\,=\,\left( \frac{8}{3}-\frac{1}{3} \right)+8(\text{In}8-\text{In}2)\,-\,7\]

\[\text{A}=\frac{7}{3}-7+16\,\text{In}\,2\]

\[\text{A}=16\,\text{In}\,2\,-\,\frac{14}{3}.\]