Question

Area bounded by \(\{(x, y) : xy \le 8 ; y \le x^2, y \ge 1\}\) is :

• A. \(16 \log_e 2 - \frac{14}{3}\)
• B. \(16 \log_e 2 - \frac{13}{3}\)
• C. \(16 \log_e 2 - \frac{17}{3}\)
• D. \(16 \log_e 2 - \frac{19}{3}\)

Hint:

Sketching the region helps identify when the upper curve changes, which indicates where the integration interval needs to be split.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The region is defined by three boundaries: the hyperbola \( y = \frac{8}{x} \), the parabola \( y = x^2 \), and the line \( y = 1 \).
We need to find the points of intersection to set up the integral.
Step 2: Key Formula or Approach:
Intersection points:
1. \( y = x^2 \) and \( y = 1 \implies x = 1 \).
2. \( y = x^2 \) and \( xy = 8 \implies x^3 = 8 \implies x = 2 \).
3. \( y = 1 \) and \( xy = 8 \implies x = 8 \).
Step 3: Detailed Explanation:
The area is divided into two parts based on the upper boundary:
Area \( = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (\frac{8}{x} - 1) dx \)
Calculation:
\[ I_1 = \left[ \frac{x^3}{3} - x \right]_1^2 = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \] \[ I_2 = \left[ 8 \ln x - x \right]_2^8 = (8 \ln 8 - 8) - (8 \ln 2 - 2) = 24 \ln 2 - 8 - 8 \ln 2 + 2 = 16 \ln 2 - 6 \] Total Area \( = \frac{4}{3} + 16 \ln 2 - 6 = 16 \ln 2 - \frac{14}{3} \).
Step 4: Final Answer:
The bounded area is \( 16 \log_e 2 - \frac{14}{3} \).