(A) 2/3 (20 log_e(2) + 9)

(B) 1/3 (40 log_e(2) + 27)

(C) 1/3 (49 log_e(2) − 15)

(D) 2/3 (24 log_e(2) − 7)

Correct Answer: 2/3 (24 log_e(2) − 7)

Explanation

The region is bounded by:

y = 1,   y = x²,   xy = 8,   x ≥ 0

From the condition xy ≤ 8, we get:

y ≤ 8/x

So for a given x, y varies from 1 to the smaller of x² and 8/x.

The curves y = x² and y = 8/x intersect when:

x² = 8/x ⇒ x³ = 8 ⇒ x = 2

Hence, area is split into two parts:

Area = ∫₁² (x² − 1) dx + ∫₂^∞ (8/x − 1) dx

Evaluating,

∫₁² (x² − 1) dx = 4/3

∫₂^∞ (8/x − 1) dx = 16 log_e(2) − 11/3

Adding both parts,

Area = 2/3 (24 log_e(2) − 7)

Hence, the required area is 2/3 (24 log_e(2) − 7).

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