The area (in sq. units) of the region bounded by the parabola \(y^2 = 8x\) and the line \(x = 2\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{32}{3}\)
The line \(x = 2\) meets the parabola where \(y^2 = 16\), that is at \(y = \pm 4\). The parabola is symmetric about the x-axis, so the area is twice the part above the x-axis, where \(y = \sqrt{8x} = 2\sqrt{2}\,\sqrt{x}\):
\( \displaystyle A = 2\int_{0}^{2} 2\sqrt{2}\,\sqrt{x}\,dx \)
\( \displaystyle = 4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_{0}^{2} \)
\( \displaystyle = 4\sqrt{2} \times \frac{2}{3} \times 2\sqrt{2} \)
\( \displaystyle = \frac{32}{3} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{3}\) — Incorrect. The area is \(\frac{32}{3}\), not \(\frac{1}{3}\).
Option (3) \(\displaystyle\frac{16}{3}\) — Incorrect. This is only the part above the x-axis. The region also has an equal part below the x-axis.
Option (4) \(\displaystyle\frac{4}{3}\) — Incorrect. The area is \(\frac{32}{3}\), not \(\frac{4}{3}\).
NCERT Exemplar:
Application of Integrals
