CUET Preparation Today
Find the area of the region bounded by the curve $y^2 = 4x$, $y$-axis and line $y = 3$. |
$2$ $\frac{9}{4}$ $\frac{9}{2}$ $\frac{15}{4}$ |
$\frac{9}{4}$ |
The correct answer is Option (2) → $\frac{9}{4}$
$\text{Required area} = \int_{0}^{3} \frac{y^2}{4} \, dy = \left[ \frac{y^3}{12} \right]_{0}^{3}$ $= \frac{27}{12} - 0 = \frac{9}{4} \text{ square units}$ |
