The area of the region enclosed between the curves \(4x^2 = y\) and \(y = 4\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{16}{3}\) sq. units
The parabola \(y = 4x^2\) meets the line y = 4 where
\(4x^2 = 4,\quad x = \pm 1\)
Between x = −1 and x = 1 the line lies above the parabola, so the area is the integral of upper curve minus lower curve:
\( \displaystyle A = \int_{-1}^{1}\left(4 - 4x^2\right)dx \)
The integrand is even, so
\( \displaystyle A = 2\int_{0}^{1}\left(4 - 4x^2\right)dx \)
\( \displaystyle = 2\left[4x - \frac{4x^3}{3}\right]_{0}^{1} \)
\( \displaystyle = 2\left(4 - \frac{4}{3}\right) = \frac{16}{3} \)
Shortcut: The area cut off from a parabola by a chord perpendicular to its axis is two-thirds of the enclosing rectangle. Here the rectangle is 2 wide (x from −1 to 1) and 4 high:
\( \displaystyle A = \frac{2}{3}\times 8 = \frac{16}{3} \)