Area of the region bounded by y = x^2 and the line y = 16 is

The correct answer is Option (1) → \(\displaystyle\frac{256}{3}\) square units

The line \(y = 16\) cuts the parabola where

    \(x^2 = 16\)

    \(x = -4 \text{ or } x = 4\)

Between these points the line lies above the parabola, so the area is the integral of (line − parabola). The region is symmetric about the y-axis, so

    \( \displaystyle \text{Area} = 2\int_{0}^{4} (16 - x^2)\,dx \)

    \( \displaystyle = 2\left[16x - \frac{x^3}{3}\right]_{0}^{4} \)

    \( \displaystyle = 2\left(64 - \frac{64}{3}\right) \)

    \( \displaystyle = 2 \times \frac{128}{3} = \frac{256}{3} \text{ square units} \)

Shortcut: A parabolic segment cut off by a chord has area \(\frac{2}{3}\) of the rectangle on that chord. Here the rectangle is 8 wide (from x = −4 to 4) and 16 high:

    \( \displaystyle \frac{2}{3} \times 8 \times 16 = \frac{256}{3} \)

NCERT Exemplar:
Application of Integrals