Question

The area of the region $\left\{\left(x,y\right):xy\le 8,1\le y\le x^2\right\}$ is

• A. $8{\log }_e\left(2\right)-\frac{14}{3}$
• B. $16{\log }_e\left(2\right)-\frac{14}{3}$
• C. $16{\log }_e\left(2\right)-6$
• D. $8{\log }_e\left(2\right)-\frac{7}{3}$

Answer 1

The correct option is B

$16{\log }_e\left(2\right)-\frac{14}{3}$

Explanation for the correct option:

Given, the equation of the curves are $\mathrm{xy}\le 8,1\le \mathrm{y}\le {\mathrm{x}}^2$

Equate the value of $\mathrm{y}$ from both the curves

$$\begin{gathered} \Rightarrow \frac{8}{\mathrm{x}}={\mathrm{x}}^2 \\ \Rightarrow \mathrm{x}=2 \end{gathered}$$

Plotting the curves, we get

The area enclosed by the curves, is the area within them.

Area described by,

$$\begin{gathered} ={\int }_1^2\left(x^2-1\right)dx+{\int }_2^8\left(\frac{8}{x}-1\right)dx \\ =16{\log }_e\left(2\right)-\frac{14}{3} \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.