Limit Definition of a Definite Integral
In this post I will talk about the limit definition of a definite integral. A crude definition of definite integral of a function is the area inside a function from from point $a$ to $b$. Suppose that we have a graph below.
The area on shade is the definite integral of $f$ from $a$ to $b$. Since we will have difficulties on calculating the area of irregular geometry, it will be much easier if we divide the area into smaller parts with $n$ partitions (we call each partition as $x$) randomly. That way, we can approximate the area using the sum of rectangles. The smaller the parts the better approximation we can get. That way, we will have graph below.
Point $a$ will have the same value as $x_{0}$ and point $b$ will have the same value as $x_{n}$. Note that the length of each partition is varying. I didn't even bother to use a scale drawing that! The length between each $x$ is called as subinterval, and it will be represented by a sampling point $c$. For example $c_{1}$ is in $[x_{0}, x_{1}]$, $c_{2}$ is in $[x_{1}, x_{2}]$, $c_{n}$ is in $[x_{n-1}, x_{n}]$. The $mesh$ of the partitions is the maximum value of all partition length $(\Delta x_{i}=x_{i}-x_{i-1})$, which is $mesh=\max_{1\leqslant i \leqslant n}(\Delta x_{i})$. In conclusion, the definite integral of $f$ on the interval $[a, b]$ can be defined as:
\[\int_{a}^{b}f(x)dx=\lim_{mesh\rightarrow 0}\sum_{i=1}^{n}f(c_{i})\Delta x_{i}\]
The smaller the mesh, the more accurate approximation we will get.
For convenience of computation let us consider that all subinterval will have the same partition length:
\[\Delta x_{i}=\frac{b-a}{n}\]
Then the subintervals can determined with:
\[c_{i}=a+(\frac{b-a}{n})i\]
With above consideration, we can alternatively define the integral of $f$ on the interval $[a, b]$ as:
\[\int_{a}^{b}f(x)dx=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(c_{i})\Delta x_{i}\]
Since the length of each partition will be the same, in order to make the $mesh$ as small as possible we need to divide the area to $n\rightarrow \infty$ partitions.
 |
| Definite Integral |
 |
| Irregular Partitions |
\[\int_{a}^{b}f(x)dx=\lim_{mesh\rightarrow 0}\sum_{i=1}^{n}f(c_{i})\Delta x_{i}\]
The smaller the mesh, the more accurate approximation we will get.
For convenience of computation let us consider that all subinterval will have the same partition length:
\[\Delta x_{i}=\frac{b-a}{n}\]
Then the subintervals can determined with:
\[c_{i}=a+(\frac{b-a}{n})i\]
 |
| Regular Partitions |
\[\int_{a}^{b}f(x)dx=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(c_{i})\Delta x_{i}\]
Since the length of each partition will be the same, in order to make the $mesh$ as small as possible we need to divide the area to $n\rightarrow \infty$ partitions.
Comments
Post a Comment