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The given equation is $100x^{2} + 25y^{2} = 2500$. Dividing both sides by $2500$, we get: $$ \frac{x^{2}}{25} + \frac{y^{2}}{100} = 1 $$ This represents an ellipse centered at the origin $(0,0)$ with $a^{2} = 25$ (so $a=5$) and $b^{2} = 100$ (so $b=10$). Since $b > a$, the major axis is along the y-axis.
To find the area using integration, we solve for $y$: $$ y^{2} = 100 \left(1 - \frac{x^{2}}{25}\right) = 4(25 - x^{2}) $$ $$ y = 2\sqrt{25 - x^{2}} $$ We consider the positive root for the upper half of the ellipse.
The area $A$ is 4 times the area in the first quadrant. The limits for $x$ are from $0$ to $5$: $$ A = 4 \int_{0}^{5} 2\sqrt{25 - x^{2}} \, dx = 8 \int_{0}^{5} \sqrt{5^{2} - x^{2}} \, dx $$
Using the standard formula $\int \sqrt{a^{2} - x^{2}} \, dx = \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$: $$ A = 8 \left[ \frac{x}{2}\sqrt{25 - x^{2}} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{5} $$ $$ A = 8 \left[ (0 + \frac{25}{2}\sin^{-1}(1)) - (0 + 0) \right] $$ $$ A = 8 \times \frac{25}{2} \times \frac{\pi}{2} = 50\pi $$
Final Answer: 50\pi \text{ square units}