The area (in sq. units) of the region \(\{(x, y): 0 \le y \le x^{2}+2,\ 0 \le y \le x+2,\ 0 \le x \le 3\}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{31}{3}\)
For each x, y runs from 0 up to the smaller of \(x^{2}+2\) and \(x+2\). Comparing them:
\((x+2)-(x^{2}+2) = x(1-x)\)
This is \(\ge 0\) for \(0 \le x \le 1\) and \(\le 0\) for \(1 \le x \le 3\). So the top of the region is \(y = x^{2}+2\) from 0 to 1 and \(y = x+2\) from 1 to 3:
\( \displaystyle A = \int_{0}^{1}(x^{2}+2)\,dx+\int_{1}^{3}(x+2)\,dx \)
\( \displaystyle \int_{0}^{1}(x^{2}+2)\,dx = \frac{1}{3}+2 = \frac{7}{3} \)
\( \displaystyle \int_{1}^{3}(x+2)\,dx = \left[\frac{x^{2}}{2}+2x\right]_{1}^{3} = \frac{21}{2}-\frac{5}{2} = 8 \)
\( \displaystyle A = \frac{7}{3}+8 = \frac{31}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{21}{2}\) — Incorrect. This is the area under y = x + 2 alone from 0 to 3, ignoring the condition \(y \le x^{2}+2\).
Option (3) 15 — Incorrect. This is the area under \(y = x^{2}+2\) alone from 0 to 3, ignoring the condition \(y \le x+2\).
Option (4) \(\displaystyle\frac{91}{6}\) — Incorrect. This takes the larger of the two curves on each part instead of the smaller one.