Question

The area of the region \[ R=\{(x,y): xy\le 8,\; 1\le y\le x^2,\; x\ge 0\} \] is:

• A. \( \dfrac{2}{3}(20\log_e 2+9) \)
• B. \( \dfrac{1}{3}(40\log_e 2+27) \)
• C. \( \dfrac{1}{3}(49\log_e 2-15) \)
• D. \( \dfrac{2}{3}(24\log_e 2-7) \)

Hint:

When a region has two possible upper boundaries, always split the integral at their point of intersection.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The region is bounded by the curves \(y = 1\), \(y = x^2\), and \(y = 8/x\).
We need to find the intersection points to set up the definite integral for the area.
Step 2: Detailed Explanation:
Intersection of \(y = x^2\) and \(y = 1\): \(x^2 = 1 \Rightarrow x = 1\).
Intersection of \(y = x^2\) and \(y = 8/x\): \(x^3 = 8 \Rightarrow x = 2\).
Intersection of \(y = 1\) and \(y = 8/x\): \(8/x = 1 \Rightarrow x = 8\).
The area is split into two parts:
1. From \(x = 1\) to \(x = 2\): Area under \(y = x^2\) and above \(y = 1\).
2. From \(x = 2\) to \(x = 8\): Area under \(y = 8/x\) and above \(y = 1\).
\[ \text{Area} = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} \left( \frac{8}{x} - 1 \right) dx \] \[ = \left[ \frac{x^3}{3} - x \right]_{1}^{2} + [8 \log_e x - x]_{2}^{8} \] \[ = \left( \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right) \right) + ( (8 \log_e 8 - 8) - (8 \log_e 2 - 2) ) \] \[ = \left( \frac{2}{3} - \left( -\frac{2}{3} \right) \right) + ( 24 \log_e 2 - 8 - 8 \log_e 2 + 2 ) \] \[ = \frac{4}{3} + 16 \log_e 2 - 6 = 16 \log_e 2 - \frac{14}{3} \] To match the options, take \(2/3\) common:
\[ \text{Area} = \frac{2}{3} (24 \log_e 2 - 7) \] Step 3: Final Answer:
The area is \(\frac{2}{3}(24 \log_e(2) - 7)\).