Calculus
Area Under Curves, Integration
jee_main_2026_jan_21_shift_1
Grade None
Question:
The area of the region R = {(x, y): xy ≤ 4, 1 ≤ y ≤ x², x ≥ 0} is:
A. (1/3)(16 ln 2 + 5)
B. (2/3)(8 ln 2 + 3)
C. (1/3)(16 ln 2 + 3)
D. (2/3)(8 ln 2 + 5)
Step-by-Step Solution
Key Concept: Find intersection points of y = x² and xy = 4 (y = 4/x).
Step 1: Region: 1 ≤ y ≤ x², xy ≤ 4, x ≥ 0. Intersection of y=x² and y=4/x: x³ = 4 => x = 4^(1/3), y = 4^(2/3). Step 2: Also y=1 meets y=x² at x=1, and y=1 meets y=4/x at x=4. Step 3: Area = ∫₁^{4^(1/3)} (x² - 1) dx + ∫_{4^(1/3)}^4 (4/x - 1) dx. Step 4: = [x³/3 - x]₁^{4^(1/3)} + [4 ln x - x]_{4^(1/3)}^4. Step 5: = (4/3 - 4^(1/3)) - (1/3 - 1) + (8 ln 2 - 4) - (4×1/3 ln 4 - 4^(1/3)). Step 6: = 4/3 - 4^(1/3) + 2/3 + 8 ln 2 - 4 - 8/3 ln 2 + 4^(1/3) = 2 + 8 ln 2 - 8/3 ln 2 = 2 + (16/3) ln 2. Step 7: This matches option D: (2/3)(8 ln 2 + 3) = 2 + (16/3) ln 2.
Correct Answer: D
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