Integral

Riemann integration

If ${\displaystyle f}$ is a real-valued function defined on an interval ${\displaystyle [a,b]}$ (where ${\displaystyle a<b}$), then the Riemann integral of ${\displaystyle f}$ over ${\displaystyle [a,b]}$ is a certain real number. It may not always be the case that we can find such a real number. If we can, the we say that ${\displaystyle f}$ is Riemann integrable over ${\displaystyle [a,b]}$.

To define the Riemann integral, we will give the following conditions for an infinite sequence:

${\displaystyle \mathbf {(i)} }$ We construct the ${\displaystyle n}$th term of the sequence as follows:

(1) Split the interval ${\displaystyle [a,b]}$ into ${\displaystyle n}$ sub-intervals ${\displaystyle [p_{0},p_{1}],[p_{1},p_{2}]...[p_{n-1},p_{n}]}$, where ${\displaystyle p_{0}=a}$, ${\displaystyle p_{n}=b}$, and ${\displaystyle p_{0}<p_{1}<p_{2}...<p_{n-1}<p_{n}}$.

(2) For each integer ${\displaystyle k}$ from ${\displaystyle 1}$ to ${\displaystyle n}$, choose an arbitrary point ${\displaystyle x_{k}}$ in the interval ${\displaystyle [p_{k-1},p_{k}]}$.

(3) Find the sum ${\displaystyle \sum _{k=1}^{n}((x_{k}-x_{k-1})f(x_{k}))}$. This is the ${\displaystyle n}$th term.

${\displaystyle \mathbf {(ii)} }$ If we are given any positive real number ${\displaystyle \delta }$, we can find a positive integer ${\displaystyle N}$ such that for all positive integers ${\displaystyle n}$ greater than ${\displaystyle N}$, the length of every sub-interval in step (1) is less than ${\displaystyle \delta }$.

If it is the case that every sequence that satisfies conditions ${\displaystyle \mathbf {(i)} }$ and ${\displaystyle \mathbf {(ii)} }$ is convergent, and converges to the same limit, only then we say that ${\displaystyle f}$ is Riemann integrable over ${\displaystyle [a,b]}$. To put it in other words, if we can find real number ${\displaystyle L}$ such that every sequence that satisfies conditions ${\displaystyle \mathbf {(i)} }$ and ${\displaystyle \mathbf {(ii)} }$ is convergent and they all converge to ${\displaystyle L}$, then ${\displaystyle f}$ is Riemann integrable over ${\displaystyle [a,b]}$, and ${\displaystyle L}$ is called the Riemann integral of ${\displaystyle f}$ over ${\displaystyle [a,b]}$.

If we can find a divergent sequence that satisfies both conditions ${\displaystyle \mathbf {(i)} }$ and ${\displaystyle \mathbf {(ii)} }$, then ${\displaystyle f}$ is not Riemann integrable over ${\displaystyle [a,b]}$. If we can find two convergent sequences that satisfy both conditions ${\displaystyle \mathbf {(i)} }$ and ${\displaystyle \mathbf {(ii)} }$, but they converge to different limits, then ${\displaystyle f}$ is not Riemann integrable over ${\displaystyle [a,b]}$.

Sum of functions

The integral of a sum of functions is the sum of each function's integral. that is,

${\displaystyle \int \limits _{a}^{b}[f(x)+g(x)]\,dx=\int \limits _{a}^{b}f(x)\,dx+\int \limits _{a}^{b}g(x)\,dx}$.

The proof of this is straightforward: The definition of an integral is a limit of sums. Thus

${\displaystyle \int \limits _{a}^{b}[f(x)+g(x)]\,dx=\lim _{n\to \infty }\sum _{i=1}^{n}\left(f(x_{i}^{*})+g(x_{i}^{*})\right)}$

${\displaystyle =\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i}^{*})+\sum _{i=1}^{n}g(x_{i}^{*})}$

${\displaystyle =\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i}^{*})+\lim _{n\to \infty }\sum _{i=1}^{n}g(x_{i}^{*})}$

${\displaystyle =\int \limits _{a}^{b}f(x)\,dx+\int \limits _{a}^{b}g(x)\,dx}$

Note that both integrals have the same limits.

Constants in integration

When a constant is in an integral with a function, the constant can be taken out. Further, when a constant c is not accompanied by a function, its value is c * x. That is,

${\displaystyle \int \limits _{a}^{b}cf(x)\,dx=c\int \limits _{a}^{b}f(x)\,dx}$ and

This can only be done with a constant.

${\displaystyle \int \limits _{a}^{b}c\,dx=c(b-a)}$

Proof is again by the definition of an integral.

Other

If a, b and c are in order (i.e. after each other on the x-axis), the integral of f(x) from point a to point b plus the integral of f(x) from point b to c equals the integral from point a to c. That is,

${\displaystyle \int \limits _{a}^{b}f(x)\,dx+\int \limits _{b}^{c}f(x)\,dx=\int \limits _{a}^{c}f(x)\,dx}$

if they are in order. (This also holds when a, b, c are not in order if we define

${\displaystyle \textstyle \int \limits _{a}^{b}f(x)\,dx=-\int \limits _{b}^{a}f(x)\,dx}$.)

${\displaystyle \int \limits _{a}^{a}f(x)\,dx=0}$

This follows the fundamental theorem of calculus (FTC): ${\displaystyle F(a)-F(a)=0}$.

${\displaystyle \int \limits _{a}^{b}f(x)\,dx=-\int \limits _{b}^{a}f(x)\,dx}$

Again, following the FTC: ${\displaystyle F(b)-F(a)=-[F(a)-F(b)]}$.