Area of the ellipse x²a² + y²b² = 1 is

The correct option is π ab sq. units.

Let us derive the area of the standard ellipse given by the equation:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

An ellipse is symmetric with respect to both the x-axis and the y-axis. Therefore, the total area $A$ of the ellipse is equal to 4 times the area of the region in the first quadrant (where $x\ge 0$ and $y\ge 0$).

First, we solve the equation of the ellipse for $y$ in terms of $x$:
$$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}$$
Multiplying both sides by $b^2$ gives:
$$y^2=b^2(1-\frac{x^2}{a^2})=\frac{b^2}{a^2}(a^2-x^2)$$
Since we are considering the first quadrant, we take the positive square root:
$$y=\frac{b}{a}\sqrt{a^2-x^2}$$

The boundary limits for $x$ in the first quadrant are from $x=0$ to $x=a$. The area in the first quadrant is given by the definite integral:
$$A_{\text{quadrant}}={\int }_0^{a}ydx={\int }_0^a\frac{b}{a}\sqrt{a^2-x^2}dx$$

The total area $A$ of the ellipse is 4 times this value:
$$A=4\cdot \frac{b}{a}{\int }_0^a\sqrt{a^2-x^2}dx$$

We use the standard integration formula:
$$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)$$

Applying the limits from 0 to $a$:
$${[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)]}_0^a$$
Evaluating this at the upper limit $x=a$:
$$\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}{\sin }^{-1}\left(1\right)=0+\frac{a^2}{2}\left(\frac{\pi }{2}\right)=\frac{\pi a^2}{4}$$
Evaluating this at the lower limit $x=0$:
$$0+\frac{a^2}{2}{\sin }^{-1}\left(0\right)=0$$

Thus, the value of the integral is $\frac{\pi a^2}{4}$. Substituting this back into our expression for the total area:
$$A=4\cdot \frac{b}{a}\left(\frac{\pi a^2}{4}\right)$$
Simplifying the constants and variables:
$$A=\pi ab$$

Therefore, the area of the ellipse is π ab sq. units.