The area (in sq. units) of the region bounded by the parabola \(y^{2} = 4(x-1)\) and the line x = 5 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{64}{3}\)
The parabola has its vertex at (1, 0) and is symmetric about the x-axis; its upper half is \(y = 2\sqrt{x-1}\). The region runs from the vertex x = 1 to the line x = 5, so the area is twice the area above the x-axis:
\( \displaystyle A = 2\int_{1}^{5}2(x-1)^{1/2}\,dx \)
\( \displaystyle = 4\left[\frac{2}{3}(x-1)^{3/2}\right]_{1}^{5} \)
\( \displaystyle = 4\times\frac{2}{3}\times8 \)
\( \displaystyle = \frac{64}{3} \)
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 = 5 meets the parabola at \(y = \pm4\), so the rectangle is 4 by 8:
\( \displaystyle A = \frac{2}{3}\times32 = \frac{64}{3} \)