Using integration the area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to :

$\pi ab $

The correct answer is Option (1) → $πab$

By symetry

$I=II=III=IV$

Required area = $4×I$

from eq.

$y=\frac{b}{a}\sqrt{a^2-x^2}$

so area = $4\frac{b}{a}\int\limits_0^a\sqrt{a^2-x^2}dx$

$=4\frac{b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a$

$=4\frac{b}{a}×\frac{πa^2}{2×2}=πab$