Find the area bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the ordinates $x = 0$ and $x = ae$, where, $b^2 = a^2(1 - e^2)$ and $e < 1$.

$ab (\sin^{-1} e + e\sqrt{1 - e^2})$

The correct answer is Option (1) → $ab (\sin^{-1} e + e\sqrt{1 - e^2})$

The required area of the region $BOB'RFSB$ is enclosed by the ellipse and the lines $x = 0$ and $x = ae$.

Note that the area of the region $BOB'RFSB$

$= 2 \int\limits_{0}^{ae} y \, dx = 2 \frac{b}{a} \int_{0}^{ae} \sqrt{a^2 - x^2} \, dx$

$= \frac{2b}{a} \left[ \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} \right]_{0}^{ae}$

$= \frac{2b}{2a} \left[ ae \sqrt{a^2 - a^2 e^2} + a^2 \sin^{-1} e \right]$

$= ab \left[ e \sqrt{1 - e^2} + \sin^{-1} e \right]$