The smaller area between the ellipse $\frac{x^2}{4}+\frac{y^2}{36}=1$ and the line $\frac{x}{2}+\frac{y}{6}=1$ is given by:

$3 \int\limits_0^2 \sqrt{4-x^2} d x-\int\limits_0^2(6-3 x) d x$ $3 \int\limits_0^2 \sqrt{4-x^2} d x+\int\limits_0^2(6-3 x) d x$ $3 \int\limits_0^4 \sqrt{4-x^2} d x-\int\limits_0^4(6-3 x) d x$ $3 \int\limits_0^4 \sqrt{4-x^2} d x+\int\limits_0^4(6-3 x) d x$

$3 \int\limits_0^2 \sqrt{4-x^2} d x-\int\limits_0^2(6-3 x) d x$

The correct answer is Option (1) → $3 \int\limits_0^2 \sqrt{4-x^2} d x-\int\limits_0^2(6-3 x) d x$