Practicing Success
Practicing Success
CUET Preparation Today
The area of region bounded by curve $y^2=4x$ and the lines $y=2$ and $x=0$ is : |
$\frac{1}{3}$ $\frac{2}{3}$ $\frac{4}{3}$ 1 |
$\frac{2}{3}$ |
The correct answer is Option (2) → $\frac{2}{3}$ $y^2=4x$, $y=2$, $x=0$
at $y=2$ $y^2=4x$ $2^2=4x$ $⇒x=1$ so required area = $2×1-\int\limits_0^12\sqrt{x}dx$ $=2-2\left[\frac{2}{3}x^{3/2}\right]_0^1$ $⇒2-\frac{4}{3}⇒\frac{2}{3}$ sq. units |
