# 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 $$\int_0^R 2\pi r\ dr = \text{Area}\ \ ?$$

## closed as unclear what you're asking by parsiad, tilper, user99914, Zain Patel, user223391 Jul 21 '16 at 4:39

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Do you mean to ask why $\int_0^i x\,dx=\frac{i^2}{2}$? – Arthur Jul 20 '16 at 17:28
• yes, to be straight to the point – Raed Tabani Jul 20 '16 at 17:32
• either, aren't you guys asking the same thing? – Raed Tabani Jul 20 '16 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 '16 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. – GeorgSaliba Jul 20 '16 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 – Andres Mejia Jul 20 '16 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 – Raed Tabani Jul 20 '16 at 18:14