The area (in square units) bounded by the curve y2 = 4x and the line x = 1 is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{8}{3}\)
The region lies between the vertex x = 0 and the line x = 1, with upper half \(y = 2\sqrt{x}\) and lower half \(y = -2\sqrt{x}\). The parabola is symmetric about the x-axis, so the area is twice the part above the x-axis:
\( \displaystyle A = 2\int_{0}^{1}2\sqrt{x}\,dx \)
\( \displaystyle = 4\left[\frac{2}{3}\,x^{3/2}\right]_{0}^{1} \)
\( \displaystyle = 4\times\frac{2}{3} = \frac{8}{3} \)
Shortcut: The area cut off from a parabola by a chord perpendicular to its axis is two-thirds of the enclosing rectangle. The line x = 1 meets the parabola at y = ±2, so the rectangle is 1 by 4:
\( \displaystyle A = \frac{2}{3}\times 4 = \frac{8}{3} \)