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So, as many of you would've been taught in middle school or high school, if you have a circle of radius r, then its circumference is 2×π×r and its area is π×r². Likely, you were simply told to memorize this, and never given any sort of explanation about how or why it was true. There is a very, very simple proof using basic calculus.

First, as above, assume we have a circle of radius r. Then we have an infinitely large set of circles of smaller size, between 0 and r. To express this in mathematical notation, the set of [0, r]. If we have a function Circ(r) = 2×π×r (basically, the circumference function), and add the result of this function for every number in this set, then we will get the area of the circle of radius r. But how are we going to do this? Integration.

First, we'll start off with the following integral:

∫₀^r 2×π×t dt
This means that we take the integral of the function 2×π×t with respect to t, and we do so from 0 to r. As 2π is a constant, it can be moved outside the integral:

2π × ∫₀^rt dt
Now we find the antiderivative of t with respect to t. (If you know what derivatives are, but not integrals, then know this. An antiderivative of a function f is a function F such that F' = f):

∫t dt = t² / 2
The indefinite integral of t is t-squared divided by two. Now we plug the indefinite integral back in, and we get:

2π × [t² / 2]₀^r
= 2π × [r² / 2 - 0² / 2]
= 2π × [r² / 2 - 0]
= 2π × [r² / 2]
= (2×π×r²) / 2
And, ultimately, the 2's cancel. This leaves us with:

π×r²
Because our initial integral was exactly equal to the area of a circle of radius r, and because our integral simplifies to this, then therefore the area of a circle of radius r is exactly equal to π×r². Q.E.D.

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NOPE.