The area bounded by $y^2=4 x$ and its latus rectum and x-axis in first quadrant is:
Answer & explanation
Correct answer: option 2
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