CUET Preparation Today
The area of the region bounded by the curves $x^2=4y$, the line $x=3$ and x-axis is : |
$\frac{3}{4}$ $\frac{7}{4}$ $\frac{9}{4}$ $\frac{18}{4}$ |
$\frac{9}{4}$ |
The correct answer is Option (3) → $\frac{9}{4}$ $x^2=4y,x=3$ so required area → $\int\limits_0^3\frac{x^2}{4}dx$ $=\left[\frac{x^3}{12}\right]_0^3=\frac{9}{4}$ sq. units
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