Area of the region bounded by the curve \(y^2 = 4x\), y-axis and the line \(y = 3\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{9}{4}\) sq. units
The region lies between the parabola and the y-axis, from y = 0 to y = 3, so it is easiest to integrate with respect to y (horizontal strips of length x):
\(\displaystyle x = \frac{y^2}{4}\)
\(\displaystyle \text{Area} = \int_{0}^{3} \frac{y^2}{4}\,dy\)
\(\displaystyle = \left[\frac{y^3}{12}\right]_{0}^{3} = \frac{27}{12} = \frac{9}{4}\) sq. units
Why the other options are incorrect
Option (1) \(\displaystyle\frac{9}{2}\) sq. units — Incorrect. This is the area on the other side of the curve: between the parabola, the x-axis and the line \(x = \frac{9}{4}\) (where y = 3), \(\int_{0}^{9/4} 2\sqrt{x}\,dx = \frac{9}{2}\). The region bounded by the y-axis is the one to the left of the curve.
Option (2) 9 sq. units — Incorrect. The area is \(\frac{9}{4}\), not 9.
Option (3) 2 sq. units — Incorrect. The area is \(\frac{9}{4}\), not 2.
NCERT (2026-27 edition):
Area under Simple Curves
