The area (in square units) of the region bounded by the curve \(x^2 = y\) and the straight line \(y = 4\) in the first quadrant is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{16}{3}\)
The parabola \(y = x^2\) meets the line \(y = 4\) where \(x^2 = 4\); in the first quadrant, x = 2. For \(0 \le x \le 2\) the line lies above the parabola, and the region is bounded on the left by the y-axis:
\(\displaystyle \text{Area} = \int_{0}^{2} (4 - x^2)\,dx\)
\(\displaystyle = \left[4x - \frac{x^3}{3}\right]_{0}^{2} = 8 - \frac{8}{3} = \frac{16}{3}\)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{2}\) — Incorrect. The area is \(\frac{16}{3}\), not \(\frac{3}{2}\).
Option (3) \(\displaystyle\frac{32}{3}\) — Incorrect. This is the area between the parabola and the line in the first and second quadrants together. The question asks only for the first quadrant, which is half of it.
Option (4) \(\displaystyle\frac{7}{2}\) — Incorrect. The area is \(\frac{16}{3}\), not \(\frac{7}{2}\).