The area of the region R = \{(x, y): x y ≤ 8, 1 ≤ y ≤ x², x ≥ 0\} is [28 Jan 2026 Shift 1]

Correct Answer is : 23(24log⁡e(2)−7)\;2/3 (24 _e(2)-7)32(24loge(2)−7)

Correct Answer is : 23(24log⁡e(2)−7)\;2/3 (24 _e(2)-7)32(24loge(2)−7)

Test Index

Area Under The Curves Part 2

Section: Mathematics

Question : 11 of 70

Marks: +1, -0

The area of the region

R = \{(x, y): x y \le 8, 1 \le y \le x^2, x \ge 0\}

Solution:

Concept:

The region is bounded by

y = 1

y = x^2

, and

y = \frac{8}{x}

in the first quadrant.

The area is found by splitting the

x

-interval where the upper boundary changes.

Explanation:

Find intersection points:

y = x^2

and

y = 1

x^2 = 1 \Rightarrow x = 1

(since

x \ge 0

y = x^2

and

y = \frac{8}{x}

x^2 = \frac{8}{x} \Rightarrow x^3 = 8 \Rightarrow x = 2

y = \frac{8}{x}

and

y = 1

\frac{8}{x} = 1 \Rightarrow x = 8

For

1 \le x \le 2

, the region is above

y = 1

and below

y = x^2

For

2 \le x \le 8

, the region is above

y = 1

and below

y = \frac{8}{x}

Area

= \int_1^2 (x^2 - 1) \, dx + \int_2^8 \left( \frac{8}{x} - 1 \right) dx

= \left[ \frac{x^3}{3} - x \right]_1^2 + \left[ 8\ln x - x \right]_2^8

= \left( \frac{8}{3} - 2 - \frac{1}{3} + 1 \right) + \left[ (8\ln 8 - 8) - (8\ln 2 - 2) \right]

= \frac{7}{3} - 1 + 8\ln 8 - 8\ln 2 - 6

= \frac{4}{3} + 8\ln(8/2) - 6 = \frac{4}{3} + 8\ln 4 - 6

= \frac{4}{3} - 6 + 16\ln 2 = -\frac{14}{3} + 16\ln 2

= \frac{2}{3} (24\ln 2 - 7)

Answer:

Option D:

\frac{2}{3} (24 \log_e(2)-7)

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