The area (in sq. units) of the region in the first quadrant bounded by the parabola \(y = x^{2}\), the line x + y = 6 and the x-axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{32}{3}\)
The parabola and the line meet where
\(x^{2} = 6-x\)
\((x+3)(x-2) = 0\)
so in the first quadrant they meet at (2, 4). The line meets the x-axis at (6, 0). The top of the region is the parabola from x = 0 to x = 2 and the line from x = 2 to x = 6, so the area is split at x = 2:
\( \displaystyle A = \int_{0}^{2}x^{2}\,dx+\int_{2}^{6}(6-x)\,dx \)
\( \displaystyle = \frac{8}{3}+\left[6x-\frac{x^{2}}{2}\right]_{2}^{6} \)
\( \displaystyle = \frac{8}{3}+(18-10) = \frac{8}{3}+8 = \frac{32}{3} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{22}{3}\) — Incorrect. This is the area between the line and the parabola from x = 0 to x = 2, which is not bounded by the x-axis.
Option (3) \(\displaystyle\frac{8}{3}\) — Incorrect. This is only the part under the parabola, from x = 0 to x = 2.
Option (4) 8 — Incorrect. This is only the part under the line, from x = 2 to x = 6.