The area (in sq. units) of the region \(\{(x,y): 3x^2 \le y \le |x|\}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{27}\)
Both \(y = 3x^2\) and \(y = |x|\) are even functions, so the region is symmetric about the y-axis. The area is twice the part with \(x \ge 0\), where \(|x| = x\).
For \(x \ge 0\) the curves meet where
\(3x^2 = x,\quad x(3x - 1) = 0,\quad x = 0 \text{ or } x = \frac{1}{3}\)
Between them \(x \ge 3x^2\) (at \(x = \frac{1}{6}\): \(\frac{1}{6} \gt \frac{1}{12}\)), so the line is on top:
\( \displaystyle \text{Area} = 2\int_{0}^{1/3} \left(x - 3x^2\right) dx \)
\( \displaystyle = 2\left[\frac{x^2}{2} - x^3\right]_{0}^{1/3} \)
\( \displaystyle = 2\left(\frac{1}{18} - \frac{1}{27}\right) \)
\( \displaystyle = 2 \times \frac{1}{54} = \frac{1}{27} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{54}\) — Incorrect. This is the area for \(x \ge 0\) only; the equal part for \(x \le 0\) is left out.
Option (2) \(\displaystyle\frac{1}{9}\) — Incorrect. This is the area under \(y = |x|\) alone from \(-\frac{1}{3}\) to \(\frac{1}{3}\); the area under the parabola is not subtracted.
Option (4) \(\displaystyle\frac{1}{36}\) — Incorrect. The area works out to \(\frac{1}{27}\), not \(\frac{1}{36}\).