notice the picture below, with an squared area, running along the ellipse, perpendicular to the x-axis
so [tex]\bf x^2+25y^2=25\implies y^2=\cfrac{25-x^2}{25}\implies y=\sqrt{1-\cfrac{x^2}{25}}
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\textit{that will be half of one side, notice in the picture}\\
\textit{so, a full side will be }2\sqrt{1-\cfrac{x^2}{25}}
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\textit{now, the area of a square is }length\times width
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but\quad
\begin{cases}
length=2\sqrt{1-\cfrac{x^2}{25}}\\\\
width=2\sqrt{1-\cfrac{x^2}{25}}
\end{cases}
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\textit{thus the area of it is }\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)[/tex]
[tex]\bf or\quad \left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2[/tex]
now... how far is it from the left-side to the right-side of the ellipse?
well, that's the major axis of it
let's see which one is that in this case
[tex]\bf x^2+25y^2=25\implies\cfrac{x^2}{25}+\cfrac{25y^2}{25}=1
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\cfrac{(x-0)^2}{5^2}+\cfrac{(y-0)^2}{1^2}=1\\\\
-----------------------------\\\\
\cfrac{(x-{{ h}})^2}{{{ a}}^2}+\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\ \ center\ ({{ h}},{{ k}})\qquad
vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\begin{cases}
a=5\\
b=1\\
h=0\\
k=0
\end{cases}\implies vertices\to (0\pm 5,0)\to (\pm 5,0)[/tex]
so the ellipse runs from -5 to 5 over the x-axis
thus the area will be [tex]\bf \displaystyle \int\limits_{-5}^{5}\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2
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\textit{or using the symmetrical property of the ellipse}
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\displaystyle 2\int\limits_{0}^{5}\left( 2\sqrt{1-\cfrac{x^2}{25}} \right)^2[/tex]
