The area (in sq. units) of the region \(\{(x, y): x^{2} \le y \le |x|+2\}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{20}{3}\)
Both \(x^{2}\) and \(|x|+2\) are unchanged when x is replaced by −x, so the region is symmetric about the y-axis. For \(x \ge 0\) the upper boundary is y = x + 2, and it meets the parabola where
\(x^{2} = x+2\)
\((x-2)(x+1) = 0,\ \ x = 2\) (taking \(x \ge 0\))
Between x = 0 and x = 2 the line is above the parabola, so the right half has area
\( \displaystyle \int_{0}^{2}(x+2-x^{2})\,dx = \left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{0}^{2} \)
\( \displaystyle = 2+4-\frac{8}{3} = \frac{10}{3} \)
Doubling for the left half:
\( \displaystyle A = 2\times\frac{10}{3} = \frac{20}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{10}{3}\) — Incorrect. This is only the right half of the region.
Option (2) \(\displaystyle\frac{9}{2}\) — Incorrect. This is the area between \(y = x^{2}\) and the line y = x + 2 (from x = −1 to x = 2), which reads |x| as x.
Option (4) 12 — Incorrect. This is the area under y = |x| + 2 from x = −2 to x = 2, without removing the part under the parabola.