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Question:

The area of the region (x,y): xy ≤ 8, 1 ≤ y ≤ x² is

16log6⁡2−1;43

16log6⁡2−6;

8log6⁡2−1;43

8log6⁡2−73

Solution:

Correct option is C. 16log6⁡2−1;43
xy ≤ 8
1 ≤ y ≤ x²
x² ≥ 1 => x ≥ 1 (since x is positive)
xy ≤ 8 => x ≤ 8/y
Require Area = ∫₁⁴ (8/y - y)dy = [8lny - y²/2]₁⁴ = 8 ln4 - 8 - (0 - 1/2) = 16ln2 - 7/2 = 16ln2 - 1/43