CUET Preparation Today
The area enclosed by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is given by : |
$3\int\limits^{4}_{0}\sqrt{9-x^2}dx$ $\frac{3}{4}\int\limits^{4}_{0}\sqrt{9-x^2}dx$ $3\int\limits^{4}_{0}\sqrt{16-x^2}dx$ $\frac{3}{4}\int\limits^{4}_{0}\sqrt{16-x^2}dx$ |
$3\int\limits^{4}_{0}\sqrt{16-x^2}dx$ |
