Riemann Integral

by ms-academy | Nov 26, 2021 | Math Learning

The story of integration started with I. Newton, when in the late 1660 he invented the method of inverse tangents to find areas under curves.

In 1680, G. Leibnitz discovered the process of finding tangent line to find area. Thus, they had discovered the integration, being a process of summation, was inverse to the operation of differentiation since finding tangent lines involved differences and finding areas involved summations.

In 1854, G.F Riemann formulated a new and different approach to define integral on the real line. He separated the concept from its differentiation. His approach was to examine the motivating summation and limit process of finding areas by itself.

In 1875, J.G Darboux viewed Riemann Integration in a different way. The approaches of both Riemann and Darboux demanded the integration to be bounded in its domain. It has been established that the two definitions of definite integral given by Riemann and Darboux are equivalent that is why Riemann integral are often called Darboux-Riemann Integrals.

In this article we will discuss about the integral based on Riemann and Darboux-Riemann approach will be discussed in mainly on real line ( $\mathbb{R}$ ).

Definition:

Partition: Let $[a, b]$ be a given closed and bounded on real line. Let $x_{0},x_{1},x_{2}, \ldots , x_{n}$ be finite number of points in $[a, b]$ such that $x_{0}=a < x_{1} < x_{2}< x_{3}<\ldots < x_{n-1}< x_{n} = b$ , then the finite order set

$P = \lbrace x_{0}, x_{1}, x_{2}, \ldots, x_{n} \rbrace$ is defined to be a partition of $[a, b]$ which divides $[a, b]$ into “ $n$ ” closed sub-interval $[ x_{i-1}, x_{i} ]$ , for all $i=1,2,\ldots, n$ .

Fact: Every partition “ $P$ ” of $[a, b]$ must include at least two points “ $a$ ” and “ $b$ ” of $[a, b]$ . In fact, $P = [a, b]$ is the trivial partition of $[a, b]$ .

Definition:

Family of partition: Every bounded and closed interval $[a, b]$ has infinite numbers of partitions. The set or the collection of all partitions of $[a, b]$ , called family of partition of $[a, b]$ , is denoted by $P [a, b]$ .Thus, if $p\in P$ $[a, b]$ implies “ $p$ ” is a partition of $[a, b]$ .

Example:

For positive integer “ $n$ ”,

$p_{1}=\left\lbrace 0,\dfrac{1}{n},\dfrac{2}{n},\cdots , \dfrac{n-1}{n} , 1 \right\rbrace$ is a partition of $[0,1]$

$p_{2}=\left\lbrace 0,\dfrac{1}{2n},\dfrac{2}{2n},\cdots , \dfrac{2n-1}{2n} , 1 \right\rbrace$ is a partition of $[0,1]$

$p_{3}=\left\lbrace 0,\dfrac{1}{3n},\dfrac{2}{3n},\cdots , \dfrac{3n-1}{3n} , 1 \right\rbrace$ is not a partition of $[0,1]$ as $1\not\in P$ .

Definition:

Norm of partition:

If $P = \lbrace x_{0}, x_{1}, x_{2}, \ldots, x_{n} \rbrace$ be a partition of $[a, b]$ and $I_{r}= [x_{r-1},\ldots, x_{r},]$ denotes the sub-intervals of $[a, b]$ (for $r=1, 2, …, n$ ) having the length $\delta r = \vert I_{r}\vert =(x_{r}-x_{r-1})$ , for $r=1, 2, \ldots, n$ .

Then the greatest length of the sub-intervals i,e., $\max \lbrace \vert I_{r} \vert \; : \; r=1, 2, \ldots, n \rbrace$ is defined to be a norm of the partition “ $p$ ” and denoted by $\delta$ or $\Vert P \Vert$ .

i,e., $\delta = max \lbrace \delta_{r} : r = 1, 2, \ldots, n\rbrace$ .

Definition:

Upper Sum: Let, $f:[a,b]\to\mathbb{R}$ be a bounded function on $[a, b]$ and $P = \lbrace x_{0}, x_{1}, x_{2}, \ldots, x_{n} \rbrace$ be a partition over $[a, b]$ .

Then “ $f$ ” is bounded on every $[ x_{r-1}, x_{r} ] (r=1, 2, \ldots, n)$ .

Let, $M_{r}=\sup\lbrace f(x) \; : \; x \in [ x_{r-1}, x_{r} ] \rbrace$

Then,
$\displaystyle\sum_{r=1}^{n}M_{r}\delta_{r}$ defined as upper sum or Darboux upper sum._r

Lower sum: Let, $f:[a,b]\to\mathbb{R}$ be a bounded function on $[a, b]$ and $P = \lbrace x_{0}, x_{1}, x_{2}, \ldots, x_{n} \rbrace$ be a partition over $[a, b]$ .

Let, $m_{r}=\inf\lbrace f(x) \; : \; x \in [ x_{r-1}, x_{r} ] \rbrace$

Then,

$\displaystyle\sum_{r=1}^{n}m_{r}\delta_{r}$ defined as lower sum or Darboux lower sum.

Notation:

Upper Sum:

For the function “ $f$ ” over the partition” $p$ ” on $[a, b]$ ,

Then,

$U(p,f)=\sum_{r=1}^{n} M_{r} \delta_{r}$

Lower Sum: For the function “ $f$ ” over the partition “ $p$ ” on $[a, b]$ ,

Then,

$L(p,f)=\sum_{r=1}^{n} m_{r} \delta_{r}$

Facts:

If $U(p,f)$ and $L(p,f)$ are equal when $f(x)$ is a constant function.

$\displaystyle U(p,f)-L(p,f)= \sum_{r=1}^{n}( M_{r}\delta_{r}-m_{r}\delta_{r})$

is the ‘oscillatory sum’ of “ $f$ ” for the partition “ $p$ ” on $[a, b]$ , and we write that as

$w(p,f)= U(p,f)-L(p,f)$

For any bounded function “ $f$ ” on $[a, b]$ and $p\in P[a,b]$ ,

$m(b-a)\leq L(p,f)\leq U(p,f)\leq M(b-a)$

Where, $m$ , $M$ are infimum and supremum of “ $f$ ” on $[a, b]$ .

Definition

Refinement of a partition:

Let, $p_{1}$ and $p_{2}$ be two partitions of $[a, b]$ such that if $p_{1}\subset p_{2} \text{ and } \delta(p_{1})\leq \delta(p_{2})$ then $p_{2}$ is called refinement of the partition “ $p_{1}$ ”.

Definition:

Common Refinement: If $p_{1}$ and $p_{2}$ are two partitions of $[a, b]$ then $p_{1}\subset p_{1}\cup p_{2}$ and $p_{2}\subset p_{1}\cup p_{2}$ .

So $p_{1}\cup p_{2}$ is the refinement of both $p_{1}$ and $p_{2}$ and is called common refinement of $p_{1}$ and $p_{2}$ .

Properties of refinement of a partition:

If “ $f$ ” is bounded on $[a, b]$ and $p_{1}, p_{2} \in P [a,b]$ where “ $p_{2}$ ” is the refinement of “ $p_{1}$ ”

Then

$\begin{align*} \text{(i).} &\; U(p_{1},f)\geq U(p_{2},f)\\ \text{(ii).} &\; L(p_{1},f)\leq L(p_{2},f) \\ \text{(iii).} &\; w(p_{1},f)\geq w(p_{2},f) \end{align*}$

Definition (Darboux Lower Integral)

Let “ $f$ ” be a real valued function over $[a,b]$ , then $\sup{\lbrace L(p,f) : p\in P [a,b]\rbrace}$ is defined as lower (Darboux) integral of “ $f$ ” on $[a,b]$ and denoted by

$\int_{\ubar{a}}^{b}f(x)dx$

Darboux Upper Integral:

For real valued function “ $f$ ” on $[a, b]$ , then $\inf{\lbrace U(p,f) : p\in P [a,b] \rbrace}$ is defined as upper (Darboux) integral of “ $f$ ” on $[a, b]$ and denoted by

$\int_{a}^{\bar{b}}f(x)dx$

Particular fact:

For any real valued bounded function on $[a, b]$ and any two partitions $p_{1}$ , $p_{2}$ of $[a, b]$

$\begin{align*} \text{(i).} &\; L(p_{1},f)\leq U(p_{2},f)\\ \text{(ii).} &\; \int_{\ubar{a}}^{b}f(x)dx \leq \int_{a}^{\bar{b}}f(x)dx\end{align*}$

Definition:

Riemann Integrability:

A real valued function “ $f$ ” on $[a, b]$ is said to be Riemann integrable on $[a, b]$ if

$\int_{\ubar{a}}^{b}f(x)dx = \int_{a}^{\bar{b}}f(x)dx$

Then “ $f$ ” is Riemann integrable and are denoted that integral by

$\int_{a}^{b}f(x)dx$

and when “ $f$ ” is Riemann integrable we denote that as $f \in \mathcal{R}[a,b]$

Example:

Let $f(x)=x,\; x\in[0,1]$

Check $f(x) \in \mathcal{R}[0,1]$

Solution: Let

$p_{n}=\left\lbrace 0,\dfrac{1}{n},\dfrac{2}{n},\cdots , \dfrac{2}{n} , 1 \right\rbrace$

be a partition of $[0, 1]$ where “ $n$ ” is a positive integer.

Then “ $p_{n}$ ” divides $[0, 1]$ into “ $n$ ” sub-intervals

$I_{r}=\left[ \dfrac{r-1}{r},\dfrac{r}{n} \right], \text{ for all } r=1,2,\ldots,n \text{ and } \delta_{r}=\dfrac{1}{n} \text{ for every } r=1,2,\ldots,n.$

Here $f$ is monotonically increasing and continuous function on $[0, 1]$

$\begin{align*} M_{r} & =\dfrac{r}{n}, \;\;\;\;\;\;\; \text{ for all } r=1,2,\ldots,n,\\ m_{r} & =\dfrac{r-1}{n}, \;\; \text{ for all } r=1,2,\ldots,n. \end{align*}$

$\begin{align*} \text{Then, } L(p_{n},f) &=\sum_{r=1}^{n}m_{r}\delta_{r}\\ & =\sum_{r=1}^{n}\dfrac{r-1}{n}\times\dfrac{1}{n}\\ & =\dfrac{1}{n^{2}}\sum_{r=1}^{n}(r-1)\\ & =\dfrac{1}{n^{2}}\times \dfrac{(n-1)n}{2}\\ L(p_{n},f) &=\dfrac{1}{2}-\dfrac{1}{2n}\\ L(p_{n},f) &<\dfrac{1}{2} \end{align*}$

Also

$\begin{align*} U(p_{n},f) &=\sum_{r=1}^{n}M_{r}\delta_{r}\\ & =\sum_{r=1}^{n}\dfrac{r}{n}\times\dfrac{1}{n}\\ & =\dfrac{1}{n^{2}}\sum_{r=1}^{n}r\\& =\dfrac{1}{n^{2}}\times \dfrac{(n+1)n}{2}\\ U(p_{n},f) &=\dfrac{1}{2}+\dfrac{1}{2n}\\ L(p_{n},f) &>\dfrac{1}{2} \end{align*}$

So,

$\sup{\lbrace L(p_{n},f): n\in \mathbb{N}\rbrace}=\dfrac{1}{2}$

$\int_{\ubar{0}}^{1}f(x)dx=\dfrac{1}{2}$

And

$\inf{\lbrace U(p_{n},f): n\in \mathbb{N}\rbrace}=\dfrac{1}{2}$

$\int_{0}^{\bar{1}}f(x)dx=\dfrac{1}{2}$

Here,

$\int_{\ubar{0}}^{1}f(x)dx=\int_{0}^{\bar{1}}f(x)dx=\dfrac{1}{2}$

Hence,

$f(x) \in \mathcal{R}[0,1]$

And

$\int_{0}^{1}f(x)dx=\dfrac{1}{2}$

So, we understand by the example how we can use the definition of Riemann Integration.

But for more complicated functions there will be difficulties to find maximum and minimum in every subinterval so, now we are going to modify the theory of “when a function is said to be a Riemann integral.”

Theorem : let $f: [a,b] \to \mathbb{R}$ be a bounded function on $[a,b]$ . A necessary and sufficient condition for inerrability of an $[a,b]$ is that for every position $\epsilon > 0 \, \exists \text{ partition } P \text{ of } [a,b]$ such that

$U(p,f)-L(p,f)<\epsilon$

Proof : (Necessary Part:)

Let $f$ be integrable on $[a,b]$ .

Then

$\int_{\ubar{a}}^{b}f(x)dx = \int_{a}^{\bar{b}}f(x)dx= \int_{a}^{b}f(x)dx$

Let $\epsilon$ be any positive number. From the definition of lower and upper integral of $f$ on $[a,b]$ , we have for the given positive $\epsilon$ , there exist a partition $P$ on $[a,b]$ such that

$L(p,f)>\int\limits_{\ubar{a}}^{b}f(x)dx-\dfrac{\epsilon}{2}=\int\limits_{a}^{b}f(x)dx-\dfrac{\epsilon}{2}$

Also

$U(p,f)<\int\limits_{a}^{\bar{b}}f(x)dx+\dfrac{\epsilon}{2}=\int\limits_{a}^{b}f(x)dx+\dfrac{\epsilon}{2}$

Therefore

$U(p,f)-L(p,f)<\int\limits_{a}^{b}f(x)dx+\dfrac{\epsilon}{2}-\int\limits_{a}^{b}f(x)dx+\dfrac{\epsilon}{2}=\epsilon$

Hence

$U(p,f)-L(p,f)<\epsilon$

Conversely (Sufficient condition)

Suppose for any positive number $\epsilon$ there exists a partition $P$ on $[a,b]$ such that

$U(p,f)-L(p,f)<\epsilon$

Since,

$\int\limits_{a}^{\bar{b}}f(x)dx\leq U(p,f)$

and

$\int\limits_{\ubar{a}}^{b}f(x)dx\geq L(p,f)$

Also

$\int\limits_{a}^{\bar{b}}f(x)dx\geq \int\limits_{\ubar{a}}^{b}f(x)dx$

Therefore

$0\leq \int\limits_{a}^{\bar{b}}f(x)dx-\int\limits_{\ubar{a}}^{b}f(x)dx\leq U(p,f)-L(p,f)<\epsilon$

Hence

Therefore

$\int\limits_{a}^{\bar{b}}f(x)dx=\int\limits_{\ubar{a}}^{b}f(x)dx$

Hence $f(x)$ is Riemann Integrable on $[a,b]$

Hence proved.

Remark:-

let $f:[a,b] \to \mathbb{R}$ be a bounded function on $[a,b]$ . If ${p_{n}}$ be a sequence if partitions of $[a,b]$ such that the sequence ${\vert p_{n}\vert}$ converges to zero and

$\lim_{n\to\infty}\lbrace U(\lbrace p_{n}\rbrace, f) - L(\lbrace p_{n}\rbrace, f)\rbrace=0$

Then $f$ is Riemann integrable on $[a,b]$ .

Now we are going to learn how to use the Remark

Example : let $f:[0,1] \to \mathbb{R}$ be a function by

$f(x)=\begin{cases} x^{2}, \; x \text{ x is rational }\\ 1, \; x \text{ is irrational} \end{cases}$

Here we are going to check the integrability of $f(x) \text{ on } [0,1]$

Lets try: Here $f(x)$ is bounded on $[0,1]$ Let for $n\in \mathbb{N}$ we choose a partition

$p_{n}=\left\lbrace 0,\dfrac{1}{n},\dfrac{2}{n},\cdots , \dfrac{2}{n} , 1 \right\rbrace$

$I_{r}=\left[ \dfrac{r-1}{r},\dfrac{r}{n} \right], \text{ for all } r=1,2,\ldots,n \text{ and } \delta_{r}=\dfrac{1}{n} \text{ for every } r=1,2,\ldots,n.$

$\text{ Let } M_{r}=\sup{\left\lbrace f(x): x\in \left[\dfrac{r-1}{n},\dfrac{r}{n} \right] \right\rbrace} \text{ for all } r=1,2,\ldots,n.$

$m_{r}=\inf{\left\lbrace f(x): x\in \left[\dfrac{r-1}{n},\dfrac{r}{n} \right] \right\rbrace} \text{ for all } r=1,2,\ldots,n.$

Here $\left[\dfrac{r-1}{n},\dfrac{r}{n} \right]$ contains rational as well as irrational points for each $r=1,2,\ldots,n.$

So,

$M_{r}=1 \text{ for } r=1,2,\ldots,n$

$m_{r}=(r-1/n)2 \text{ for } r=1,2,\ldots,n.$

Then,

$\begin{align*} U(p_{n},f) &=\sum_{r=1}^{n}M_{r}\delta_{r}\\ & =\dfrac{1}{n}\sum_{r=1}^{n}1\\ U(p_{n},f) &=1 \end{align*}$

Also

$\begin{align*} L(p_{n},f) &=\sum_{r=1}^{n}m_{r}\delta_{r}\\ & =\dfrac{1}{n^{3}}\sum_{r=1}^{n}(r-1)^{2}\\ & =\dfrac{(n-1)(2n-1)}{6n^{2}} \end{align*}$

Then

$\lim_{n\to\infty}\lbrace U(p_{n},f)-L(p_{n},f) \rbrace=1-\dfrac{1}{3}=\dfrac{2}{3}\neq 0.$

Hence by the Remark we can say that $f$ is not Riemann integral on $[0,1]$ .

Properties of Riemann integrable functions :

If $f \, : \, [a,b]\to \mathbb{R}$ is monotonic then $f \in \mathbb{R} [a,b]$ . i.e $f$ is Riemann integrable.

If $f \, : \, [a,b]\to \mathbb{R}$ is continuous then $f$ is Riemann integrable.

If $f \, : \, [a,b]\to \mathbb{R}$ be bounded but has finite number of discontinuous points on $[a,b]$ then $f \in \mathbb{R} [a,b]$

If $f \, : \, [a,b]\to \mathbb{R}$ bounded but has infinite point on discontinuities in $[a,b]$ such that the number of limits Of these infinite discontinuous points is finite in $[a,b]$ then $f(x)\in \mathcal{R}[0,1]$ .

If $f \, : \, [a,b]\to \mathbb{R}$ is integrable on $[a,b]$ , then $f$ is integrable on every closed sub interval of $[a,b]$ .

If $a<c<b$ , and $f \, : \, [a,b]\to \mathbb{R}$ is integrable on $[a,c]$ and on $[a,b]$ then

$\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=\int_{a}^{b}f(x)dx$

If $f \, : \, [a,b]\to \mathbb{R}$ is integrable on $[a,b]$ then for every $\lambda\in\mathbb{R}$ then $\lambda f(x)$ is also integrable on $[a,b]$ and further more,

$\int_{a}^{b}\lambda f(x)dx=\lambda\int_{a}^{b}f(x)dx$

If $f; g \, : \, [a,b]\to \mathbb{R}$ two integrable on $[a,b]$ and

If $f \, : \, [a,b]\to \mathbb{R}$ and $g \, : \, [a,b]\to \mathbb{R}$ then for any $\lambda,\mu \in \mathbb{R} \; \lambda f(x)+\mu g(x)$ integrable on $[a,b]$ furthermore,

$\int_{a}^{b}\lbrace \lambda f(x)+\mu g(x)\rbrace dx=\lambda\int_{a}^{b}f(x)dx+\mu\int_{a}^{b}g(x)dx$

(x) If $f \, : \, [a,b]\to \mathbb{R}$ is integrable on $[a,b]$ , then $\vert f(x) \vert$ on $[a,b]$ integrable. But the converse is not true.

(Why?)

$\text{Consider }f(x)=\begin{cases} \;\;\; 1, \; x \text{ x is rational }\\ -1, \; x \text{ is irrational} \end{cases}$

Here, $f(x)$ is not Riemann integrable but

$\vert f(x)\vert= 1\;\; \forall x \in [a,b]$

$\vert f(x)\vert$ is constant then it must be Riemann integrable on $[a,b]$ .

If $f \, : \, [a,b]\to \mathbb{R}$ and $g \, : \, [a,b]\to \mathbb{R}$ are both integrable on $[a,b]$ , then $fg$ also integrable on $[a,b]$ .

If $f \, : \, [a,b]\to \mathbb{R}$ is integrable on $[a,b]$ , these $f^{2}$ is integrable on $[a,b]$ but the converse is not true (try with the same function

$f(x)=\begin{cases} \;\;\; 1, \; x \text{ x is rational }\\ -1, \; x \text{ is irrational} \end{cases}$

If $f \, : \, [a,b]\to \mathbb{R}$ and $g \, : \, [a,b]\to \mathbb{R}$ integrable and $g(x) \geq k \; \forall x\in [a,b]$ where $k>0$ , then $f/g$ also integrable as $[a,b]$ .

If $f \, : \, [a,b]\to \mathbb{R}$ is Riemann integrable on $[a,b]$ and $f(x) \geq k \; \forall x\in [a,b]$ where $k>0$ then $1/f$ also integrable on $[a,b]$ .

If $I=[a,b]$ and $J=[c,d]$ bounded and closed interval. Let $f \, : \, I\to \mathbb{R}$ and $f \, : \, J\to \mathbb{R}$ integrable and continuous function such that $f(I) \subset [c,d]$ then $g\circ f : I\to \mathbb{R}$ is integrable on $I$ .

Example 1 : – let $f:[0,1]\to \mathbb{R}$ be a bounded function on $[0,1]$ defined by

$f(x)=\begin{cases} 1, \;\;\;\;\;\;\;\; x =0\\ x, \; 0<x\leq 1 \end{cases}$

Now we are going to check $\dfrac{1}{f(x)}$ is Riemann Integrable or not.

Let’s start:- Here $f:[0,1]\to \mathbb{R}$ is bounded function also $f$ is continuous on $[0,1]$ except $x=0$ , so $f$ is Riemann Integrable on $[0,1]$ .

But

$\dfrac{1}{f(x)}=\begin{cases} 1, \;\;\;\;\;\;\;\; x =0\\ \dfrac{1}{x}, \; 0<x\leq 1 \end{cases}$

As $x\to 0+$ , $\dfrac{1}{f(x)}\to\infty +$ , so $\dfrac{1}{f(x)}$ is not bounded on $[0,1]$ and hence $\dfrac{1}{f(x)}$ is not Riemann Integrable on $[0,1]$ and we are done.

Example 2 Consider the function $f(x):[0,4]\to \mathbb{R}$ such that

$f(x)=x-[x]$

Here also we are going to check $f(x)$ integrable or not in $[0,4]$ and if integrable then we will find the value of the integration.

Let’s start

Here the expression of $f(x)$ is

$f(x)=\begin{cases} x,& \;\; 0\leq x< 1\\ x-1,& \; 1\leq x<2\\ x-2,& 2\leq x <3\\ x-3,& 3\leq x <4\\ 0,& x=4 \end{cases}$

Since $0\leq f(x)<1$ for all $x$ in $[0,4]$ , $f(x)$ is bounded on $[0,4]$ . $f(x)$ is continuous on $[0,4]$ except the points $x=1, x=2, x=3,$ and $x=4$ . So $f(x)$ has finite number of discontinuity on $[0,4]$ then we can say that $f$ is integrable on $[0,4]$

So hence $f(x)$ is Riemann Integrable on $[0,4]$

Now we are going to find the value of the integration of the function $f(x)$ on $[0,4]$ .

Let we define

$\begin{align*} g_{1}(x) &=x \text{ on } 0\leq x <1\\ g_{2}(x) &=x-1 \text{ on } 1\leq x <2\\ g_{3}(x) &=x-2 \text{ on } 2\leq x <3\\ g_{4}(x) &=x-3 \text{ on } 3\leq x <4\\ \end{align*}$

And then $f(x)$ becomes

$f(x)=\begin{cases} g_{1}(x),& \;\; 0\leq x< 1\\ g_{2}(x),& \; 1\leq x<2\\ g_{3}(x),& 2\leq x <3\\ g_{4}(x),& 3\leq x <4\\ 0,& x=4 \end{cases}$

Hence

$\int_{0}^{4}f(x)dx=\int_{0}^{1}f(x)dx+\int_{1}^{2} f(x)dx+\int_{2}^{3} f(x)dx+\int_{3}^{4} f(x)dx$

Therefore we can do this now

$\begin{align*} \int_{0}^{4}f(x)dx & =\int_{0}^{1}g_{1}(x)dx+\int_{1}^{2} g_{2}(x)dx+\int_{2}^{3} g_{3}(x)dx+\int_{3}^{4} g_{4}(x)dx\\ & =\int_{0}^{1}xdx+\int_{1}^{2}(x-1)dx+\int_{2}^{3}(x-2)dx+\int_{3}^{4}(x-3)dx\\ & =\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=2 \end{align*}$

Therefore,

$\int_{0}^{4}f(x)dx=2$

And we are done with this problem

Now try this problem with same manner $f(x):[0,4]\to \mathbb{R}$ such that

$f(x)=x[x]$

Try to prove $f(x)$ is Riemann Integrable and find $\displaystyle\int_{0}^{4}f(x)dx$ (Answer $\displaystyle\int_{0}^{4}f(x)dx=17$ )

Some Inequalities related with Riemann Integration:

If $f:[a,b]\to \mathbb{R}$ is integrable on $[a, b]$ and $f(x)\geq 0$ for all values of $x$ in $[0, 1]$ .

Then

$\int_{a}^{b}f(x)dx\geq 0$

If $f:[a,b]\to \mathbb{R}$ and $g : [a,b]\to \mathbb{R}$ is integrable on $[a, b]$ and $f(x)\geq g(x)$ for all values of $x$ in $[0, 1]$ .

Then

$\int_{a}^{b}f(x)dx\geq \int_{a}^{b}g(x)dx$

If $f : [a,b]\to \mathbb{R}$ is integrable on $[a, b]$

Then

$\left\vert\int_{a}^{b}f(x)dx\right\vert\leq \int_{a}^{b}\vert f(x)\vert dx$

Problems related with these theories:-

Example.

Show that

$\dfrac{1}{2}\leq\int_{0}^{1}\dfrac{dx}{\sqrt{4-x^{2}+x^{3}}}\leq \dfrac{\pi}{6}$

To prove this we are going to use all of the theories which we learnt till now,

Let

$g(x)=\dfrac{1}{\sqrt{4-x^{2}+x^{3}}}, x\in[0,1]$

Here for all $x\in[0,1]$

$0\leq x^{3}\leq x^{2} \leq 1$

For all $x\in[0,1]$

$0\leq 4-x^{2}\leq 4- x^{2}+ x^{3}\leq 4,$

For all $x\in[0,1]$

$\dfrac{1}{2}\leq\dfrac{1}{\sqrt{4-x^{2}+x^{3}}}\leq \dfrac{1}{\sqrt{4-x^{2}}}$

Therefore let for all $x\in[0,1]$

$f(x)=\dfrac{1}{\sqrt{4-x^{2}}}$

and

$h(x)=\dfrac{1}{2}$

Here

for all $x\in[0,1]$

$f(x)\geq g(x) \geq h(x)$

So therefore

$\int_{0}^{1}f(x)dx\geq \int_{0}^{1}g(x)dx\geq \int_{0}^{1}h(x)dx$

$\int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}}}dx\geq \int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}+x^{3}}}dx\geq \int_{0}^{1}\dfrac{1}{2}dx$

$\int_{0}^{1}\dfrac{1}{2}dx\leq \int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}+x^{3}}}dx\leq\int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}}}dx$

Now

$\int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}}}dx=\left[\sin^{-1}\left(\dfrac{x}{2}\right)\right]_{0}^{1}$

$\int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}}}dx=\dfrac{\pi}{6}$

And

$\int_{0}^{1}\dfrac{1}{2}dx=\dfrac{1}{2}$

Therefore

$\int_{0}^{1}\dfrac{1}{2}dx\leq \int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}+x^{3}}}dx\leq\int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}}}dx$

$\dfrac{1}{2}\leq \int_{0}^{1}\dfrac{1}{\sqrt{4-x^{2}+x^{3}}}dx\leq\dfrac{\pi}{6}$

Hence proved.

Now some work sheet problems:-

Let
$f(x)=x^{2}, x\in[0,1]$

Check $f(x)$ is Riemann Integrable or not.

Try to prove these problems

$\dfrac{1}{3\sqrt{2}}\leq \int_{0}^{1}\dfrac{x^{2}dx}{\sqrt{1+x}}\leq \dfrac{1}{3}$

$\dfrac{1}{2}\left(1-\dfrac{1}{e}\right)\leq \int_{0}^{1}e^{-x^{2}}dx\leq 1$

iii)

$\dfrac{1}{3}\leq \int_{0}^{1}\dfrac{x^{2}dx}{\sqrt{1+x+x^{2}}}\leq \dfrac{\pi}{4}$

Try to show that

$\int_{0}^{\frac{\pi}{2}}\sin^{n+1}xdx\leq \int_{0}^{\frac{\pi}{2}}\sin^{n} xdx$

Bibliography

Elementary Analysis: The theory of Calculus; Kenneth Ross.
Improper Riemann Integration; Markos Roussos
Modern Theories of Integration; H. Kestelman
The Riemann Approach to Integration; Washek Pfeffer.

Thank you………

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Riemann Integral

Definition (Darboux Lower Integral)

Riemann Integrability:

Properties of Riemann integrable functions :

Some Inequalities related with Riemann Integration:

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