Question

The area of the region given by \(\left\{(x, y): x y \leq 8,1 \leq y \leq x^2\right\}\) is :

• A. \(16 \log _e 2-\frac{14}{3}\)
• B. \(8 \log _e 2-\frac{13}{3}\)
• C. \(8 \log _e 2+\frac{7}{6}\)
• D. \(16 \log _e 2+\frac{7}{3}\)

Hint:

When calculating the area of a region defined by inequalities, carefully analyze the boundaries and split the integral into parts based on the intervals of the variables.

Answer 1

The Correct Option is A

The region is defined by:
\[xy \leq 8, \quad 1 \leq y \leq x^2.\]
The boundaries of the region are: \(y = 1\), \(y = x^2\), \(y = \frac{8}{x}\).
The region of integration is split into two parts:
For \(x \in [1, 2]\), the area is bounded by \(y = x^2\) and \(y = 1\).
For \(x \in [2, 8]\), the area is bounded by \(y = \frac{8}{x}\) and \(y = 1\).
Step 1: Set up the Area Integral
The total area is:
\[\text{Area} = \int_1^2 (x^2 - 1) \, dx + \int_2^8 \left(\frac{8}{x} - 1\right) \, dx.\]
Step 2: Evaluate the First Integral
\[\int_1^2 (x^2 - 1) \, dx = \int_1^2 x^2 \, dx - \int_1^2 1 \, dx.\]
\[\int_1^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.\]
\[\int_1^2 1 \, dx = \left[x\right]_1^2 = 2 - 1 = 1.\]
\[\int_1^2 (x^2 - 1) \, dx = \frac{7}{3} - 1 = \frac{4}{3}.\]
Step 3: Evaluate the Second Integral
\[\int_2^8 \left(\frac{8}{x} - 1\right) \, dx = \int_2^8 \frac{8}{x} \, dx - \int_2^8 1 \, dx.\]
\[\int_2^8 \frac{8}{x} \, dx = 8 \int_2^8 \frac{1}{x} \, dx = 8[\ln x]_2^8 = 8(\ln 8 - \ln 2) = 8\ln 4 = 16\ln 2.\]
\[\int_2^8 1 \, dx = \left[x\right]_2^8 = 8 - 2 = 6.\]
\[\int_2^8 \left(\frac{8}{x} - 1\right) \, dx = 16\ln 2 - 6.\]
Step 4: Add Both Integrals
\[\text{Area} = \frac{4}{3} + (16\ln 2 - 6).\]
\[\text{Area} = 16\ln 2 - 6 + \frac{4}{3} = 16\ln 2 - \frac{18}{3} + \frac{4}{3} = 16\ln 2 - \frac{14}{3}.\]
Conclusion: The area of the region is \(16\ln 2 - \frac{14}{3}\)