# how is it that $\int_0^x 2\pi r\ dr$ is equal to the area of a circle [closed]

I'm studying calculus and I'm having some basic questions, this one is regarding the area of a circle. we know, from some guy, that the circumference of a circle is $2 \pi r$ and the area can be seen as the sum of all the circumferences from $0$ to $r$ which is the same as the integral, which leads us to $\pi r^2$. my question is how is it that $$\int_0^R 2\pi r\ dr = \text{Area}\ \ ?$$

• Do you mean to ask why $\int_0^i x\,dx=\frac{i^2}{2}$? Jul 20, 2016 at 17:28
• yes, to be straight to the point Jul 20, 2016 at 17:32
• either, aren't you guys asking the same thing? Jul 20, 2016 at 17:36
• @RaedTabani I edited your question. Am I right in thinking that this is what you meant? If not, you can rollback the edit by clicking "edited [X time] ago" on your question and clicking "rollback" on the previous version.
– user137731
Jul 20, 2016 at 17:45
• @Bye_World I believe that the OP's original question was how the sum of a certain variable $k$ from $k=0$ to $n$ is $n^2/2$ and not $n(n+1)/2$ for example. His issue I think is with stating the sentence "is a sum of ..." mathematically which led him to think that it's the sum of $k$ and not the Riemann sum. Jul 20, 2016 at 17:58

Here is a proof with few words. Note that the line is the graph of $y=t$ with $t$ being the horizontal axis and that the integral

$$\int_0^x t\,dt$$ represents the area of a triangle with both base and height equal to $x$.

But for more general functions, one needs the Riemann Sum as indicated by GeorgSaliba.

• This is wonderful! I never thought to explain it this way, although I suspect it would be difficult to extend this "proof" to higher degree polynomials Jul 20, 2016 at 17:55

$$\int_0^R xdx\ne\sum_0^Ri$$

You are adding small areas of circumference equal to $2\pi r$ and thickness $\Delta r$ over a span of $r=0$ to $R$. So you have to partition this interval into $n$ subdivisions, each subdivision is $R/n$ thick. This means that each radius to be added can be expressed like this $$r={i\times R\over n}\qquad i=0,1,2,\dots,n$$and $$\Delta r=\frac Rn$$ When $n$ is "infinitely large" you get an "infinitely precise" sum, which is then: $$\int_0^R2\pi rdr=\lim_{n\rightarrow\infty}2\pi\frac Rn\sum_0^n\frac{Ri}{n}=\lim_{n\rightarrow\infty}2\pi\frac{R^2}{n^2}\frac {n(n+1)}{2}=\lim_{n\rightarrow\infty}2\pi\frac{R^2}{n^2}\frac {n\times n}{2}=\pi R^2$$

This is called a Riemann Sum

• thank you this is exactly what I was looking for Jul 20, 2016 at 18:14