The area (in sq. units) of the region bounded by the parabola \(y^2 = 4x\) and the line \(x = 1\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{8}{3}\)
The parabola \(y^2 = 4x\) is symmetric about the x-axis, and its upper half is \(y = 2\sqrt{x}\). The region lies between x = 0 and x = 1, so its area is twice the area of the upper half:
\( \displaystyle A = 2\int_{0}^{1} 2\sqrt{x}\,dx \)
\( \displaystyle A = 4\left[\frac{2}{3}x^{3/2}\right]_{0}^{1} = \frac{8}{3} \)
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 1 wide (x from 0 to 1) and 4 high (y from −2 to 2), so the area is \(\frac{2}{3} \times 4 = \frac{8}{3}\).
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{3}\) — Incorrect. The area is \(\frac{8}{3}\), not \(\frac{1}{3}\).
Option (2) \(\displaystyle\frac{4}{3}\) — Incorrect. This is the area of the upper half only (above the x-axis); the region also has an equal part below the x-axis.
Option (3) \(\displaystyle\frac{5}{3}\) — Incorrect. The area is \(\frac{8}{3}\), not \(\frac{5}{3}\).