The area of the region bounded by curves y=x^2 and line y=4 is:

The correct answer is Option (3) → \( \displaystyle\frac{32}{3} \) sq. units

The curves meet where \( x^{2}=4 \), i.e. x = −2 and x = 2. Between them the line y = 4 lies above the parabola:

    \( \displaystyle \text{Area}=\int_{-2}^{2}(4-x^{2})\,dx \)

    \( \displaystyle =2\int_{0}^{2}(4-x^{2})\,dx \)

    \( \displaystyle =2\left[4x-\frac{x^{3}}{3}\right]_{0}^{2} \)

    \( \displaystyle =2\left(8-\frac{8}{3}\right) \)

    \( \displaystyle =\frac{32}{3}\text{ sq. units} \)

Why the other options are incorrect

• Option (1) \( \displaystyle\frac{48}{3} \) sq. units — Incorrect. \( \frac{48}{3}=16 \) is the whole rectangle 4 × 4, without taking away the part under the parabola.

• Option (2) \( \displaystyle\frac{16}{3} \) sq. units — Incorrect. \( \frac{16}{3} \) is only the right half (0 ≤ x ≤ 2); the region is symmetric about the y-axis and has two such halves.

• Option (4) \( \displaystyle\frac{8}{3} \) sq. units — Incorrect. \( \frac{8}{3}=\int_{0}^{2}x^{2}\,dx \) is the area under the parabola, outside the region.

Tip: The region between \( y=x^{2} \) and y = c is two-thirds of the enclosing rectangle: here \( \frac{2}{3}\times 16=\frac{32}{3} \).