CUET Preparation Today
The area bounded by $y^2=4 x$ and its latus rectum and x-axis in first quadrant is: |
0 sq. units $\frac{4}{3}$ sq. units $\frac{2}{3}$ sq. units $\frac{1}{3}$ sq. units |
$\frac{4}{3}$ sq. units |
The correct answer is Option (2) - $\frac{4}{3}$ sq. units
so area = $\int\limits_0^1ydx=\int\limits_0^1\sqrt{4x}dx$ $=2\frac{2}{3}[x\sqrt{x}]_0^1$ $=\frac{4}{3}$ sq. units |
