A beginners' proof for area under curve

BS"D

Hi all :)

Just starting to learn Calculus and we're at limits/derivatives.

Something I always wanted to do but never had the knowledge to is finding areas under curves. People always told me calculus teaches that, but I really don't know how long it'll take until we get there, so I thought I'll use what I currently know and try and find the area under $y=x^2$, between the origin and the point $(4,0)$.

After reading online that calculus uses infinitly thin rectangles and summs their areas, I set to do so myself.

Long story short, I found that if you calculate the area in two ways, one way where the origin is your starting point and another way where the second point $(4,0)$ is your starting point, and then compare both answers, all infinities cancel out and you're left with a simple subtraction where both operands seem to be the derivatives of the original function.

I have two questions, if I may:

1. The way I did things is by alternating between the two points as the starting point and then comparing the results. Is this the way mathematicians use to calculate the area under curve? I'm sure there is a generalization, but when first figuring things out, is this the way it was done?

2. I also read online that there is a link between limits\derivative and areas under curves. So far, I could not find the connection between the two. It definitely looks like the derivitives of the function are there, but other than calling it a coincidince, I couldn't find a mathematical way (that is, using algebraic methods) of going from limits\derivatives to the area under a curve. Am I missing something?

Thanks a lot in advance,

Hoping my questions were clear :)

• Arye Segal.
• BS"D I thank everyone who replied. Nice to see how it was/is done :) Have a great week. – Arye Segal Sep 10 '17 at 12:14

I can answer both questions at once. The way mathematicians compute that area consists in finding a function $f\colon[0,4]\longrightarrow\mathbb R$ such that $f'(x)=x^2$. In this case, you can take $f(x)=\frac{x^3}3$. Then the area that you are interested in is $f(4)-f(0)=\frac{64}3$.

The way mathematicians did this centuries ago, before the creation of Calculus, was to divide the interval $[0,4]$ in intervals of the same length ($[a_0,a_1]\cup[a_1,a_2]\cup\ldots\cup[a_{n-1},a_n]$) and to compute the areas of the rectangles thus obtained. By this I mean two kinds of rectangles:

• those whose vertices were $(a_k,0)$, $(a_{k+1},0)$, $(a_{k+1},{a_{k+1}}^2)$, and $(a_k,{a_{k+1}}^2)$;
• those whose vertices were $(a_k,0)$, $(a_{k+1},0)$, $(a_{k+1},{a_k}^2)$, and $(a_k,{a_k}^2)$.

The rectangles of the first type form a region whose area is greater than the area of the area of the region that you are interested in, whereas the rectangles of the first type form a region whose area is smaller than the area of the area of the region that you are interested in. Then, if, as the number of intervals increases, these areas approach the same value, that value would be the area of your region.

The connection you are looking for is known as the Fundamental Theorem of Calculus.

It says that if the area under the curve of equation $y=f(x)$, between the vertical limits $0$ and $x$ is the function $A(x)$, then $f$ is the derivative of $A$. For this reason we say that $A$ is the antiderivative of $f$.

The geometric explanation is easy: the difference of areas between $x$ and $x+h$, i.e. $A(x+h)-A(x)$ is a single rectangle of area $hf(x)$, hence

$$f(x)\approx\frac{A(x+h)-A(x)}h\approx A'(x).$$

So in practive, you don't play with a sum of tiny rectangles but look for antiderivatives instead.

For your parabola, $f(x)=x^2$ we have

$$A'(x):=\left(\frac{x^3}3\right)'=x^2$$ and the area is given by

$$A(x)=\frac{x^3}3.$$

You will sooner or later learn how to find the antiderivative of a function. And you will also see that this cannot always be done using the usual functions and new ones must be invented.

An important case is that of the area of the circle. Unfortunately, it is a difficult one. For the right half of a unit circle, you have (check by taking the derivative)

$$A(x)=2\int\sqrt{1-x^2}dx=x\sqrt{1-x^2}+\arcsin x$$

giving, for a full circle

$$2A(1)=0+2\arcsin(1)=\pi.$$