1. **Problem statement:** Solve the integral of the ellipse equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with respect to $x$ or $y$ within given limits. 2. **Understanding the ellipse equation:** The ellipse equation can be rearranged to express $y$ in terms of $x$: $$y = \pm b \sqrt{1 - \frac{x^2}{a^2}}$$ 3. **Integral setup:** To find the area under the ellipse curve from $x = -a$ to $x = a$, we integrate the positive half of $y$ and multiply by 2: $$\text{Area} = 2 \int_{-a}^a b \sqrt{1 - \frac{x^2}{a^2}} \, dx$$ 4. **Substitution:** Let $u = \frac{x}{a}$, so $x = au$ and $dx = a \, du$. The limits change from $x = -a$ to $x = a$ into $u = -1$ to $u = 1$: $$\text{Area} = 2b \int_{-1}^1 a \sqrt{1 - u^2} \, du = 2ab \int_{-1}^1 \sqrt{1 - u^2} \, du$$ 5. **Integral evaluation:** The integral $\int_{-1}^1 \sqrt{1 - u^2} \, du$ represents the area of a semicircle of radius 1, which equals $\frac{\pi}{2}$. 6. **Final area:** Substitute the integral result: $$\text{Area} = 2ab \times \frac{\pi}{2} = \pi ab$$ **Answer:** The area enclosed by the ellipse is $$\boxed{\pi ab}$$.