The area (in sq. units) of the region \(\{(x, y): x^{2} \le y \le 4,\ x \ge 1\}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5}{3}\)
The condition \(x^{2} \le 4\) gives \(-2 \le x \le 2\), and with \(x \ge 1\) the region runs from x = 1 to x = 2. For each such x, y goes from the parabola up to the line y = 4, so the width is \(4-x^{2}\):
\( \displaystyle A = \int_{1}^{2}(4-x^{2})\,dx \)
\( \displaystyle = \left[4x-\frac{x^{3}}{3}\right]_{1}^{2} \)
\( \displaystyle = \left(8-\frac{8}{3}\right)-\left(4-\frac{1}{3}\right) \)
\( \displaystyle = \frac{16}{3}-\frac{11}{3} = \frac{5}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{7}{3}\) — Incorrect. This is \(\displaystyle\int_{1}^{2}x^{2}\,dx\), the area below the parabola; the region lies above the parabola.
Option (2) \(\displaystyle\frac{16}{3}\) — Incorrect. This is the area from x = 0 to x = 2; the strip from x = 0 to x = 1 is not in the region.
Option (3) \(\displaystyle\frac{11}{3}\) — Incorrect. This is the strip from x = 0 to x = 1, the part left out by the condition \(x \ge 1\).