I was studying determination of area in Calculus. So, I decided to calculate area of rectangle using Calculus.

Let, length of a line of a square is $5$. So, I decided to make an equation for that. I took $x^2+y^2=\sqrt{25}$. Firstly, it was looking perfect to me. Then, when I was calculating for $y=1$ then, I noticed I got $x=4,6$ (where 6 is wrong).

I drew the above picture (Sorry! I don't have proper "picture" maker in system)..

We know $$\int_{-5}^{5}y\mathrm dx$$ $$=\int_{-5}^{5} \sqrt{\sqrt{25}-x^2}\mathrm dx$$

If I integrate it than, I get imaginary number and infinite number. So, I think I took the wrong equation. How would you determine area of square using Integration (when length of a line is $5$)?

Note : Use Cartesian Coordinate.

Ohh! Sorry. I wrote $x^2+y^2=\sqrt{25}$ but, when doing that in notepad I was doing using $x+y=\sqrt{25}$ then, I got $50$...

$$\int_{-5}^{5} \sqrt{25}-x \mathrm dx$$ $$[\sqrt{25}x]-\frac{x^2}{2}]_{-5}^5$$ $$\sqrt{25}(5)-\sqrt{25}(-5)-\frac{25}{2}+\frac{25}{2}$$ $$2\sqrt{25}(5)$$ $$50,-50$$