Riemann Integral

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

Table of Contents

Introduction:

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 $x_{0}=a < x_{1} < x_{2}< x_{3}$ 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] $. <ul> <li>Definition: </li> </ul> Family of partition: Every bounded and closed interval$ [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] $.Thus, if$ pin P[a, b] $implies ``$ p $'' is a partition of$ [a, b] $. Example: For positive integer ``$ n $'',$ p_{1}=leftlbrace 0,dfrac{1}{n},dfrac{2}{n},cdots , dfrac{n-1}{n} , 1 rightrbrace $is a partition of$ [0,1]p_{2}=leftlbrace 0,dfrac{1}{2n},dfrac{2}{2n},cdots , dfrac{2n-1}{2n} , 1 rightrbrace $is a partition of$ [0,1]p_{3}=leftlbrace 0,dfrac{1}{3n},dfrac{2}{3n},cdots , dfrac{3n-1}{3n} , 1 rightrbrace $is not a partition of$ [0,1] $as$ 1notin P $. <ul> <li>Definition:</li> </ul> Norm of partition: If$ 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, nrbrace $. <ul> <li>Definition:</li> </ul> Upper Sum: Let,$ f:[a,b]tomathbb{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}=suplbrace f(x) ; : ; x in [ x_{r-1}, x_{r} ] rbrace $Then,$ displaystylesum_{r=1}^{n}M_{r}delta_{r} $defined as upper sum or Darboux upper sum. r Lower sum: Let,$ f:[a,b]tomathbb{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}=inflbrace f(x) ; : ; x in [ x_{r-1}, x_{r} ] rbrace $Then,$ displaystylesum_{r=1}^{n}m_{r}delta_{r} $defined as lower sum or Darboux lower sum. r Notation: Upper Sum: For the function ``$ f $'' over the partition''$ p $'' on$ [a, b]

$, Then, <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-e6e8086e2900853be37438a123232fee_l3.png" height="26" width="295" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[U(p,f)=sum_{r=1}^{n} M_{r} delta_{r}\]" title="Rendered by QuickLaTeX.com"/> Lower Sum: For the function ``$

f $'' over the partition ``$ p $'' on$ [a, b]

$, Then, <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-a4455c31af7e203773cabd3580051fd9_l3.png" height="26" width="291" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[L(p,f)=sum_{r=1}^{n} m_{r} delta_{r}\]" title="Rendered by QuickLaTeX.com"/> Facts: <ol> <li>If$

U(p,f) $and$ L(p,f) $are equal when$ f(x)

$is a constant function.</li> </ol> 2. <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-c7f0357709f1b56775c36d5ac8fa676d_l3.png" height="26" width="687" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[displaystyle U(p,f)-L(p,f)= sum_{r=1}^{n}( M_{r}delta_{r}-m_{r}delta_{r})\]" title="Rendered by QuickLaTeX.com"/> is the 'oscillatory sum' of ``$

f $'' for the partition ``$ p $'' on$ [a, b]

$, and we write that as <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-7e6582ca65984bd1addd6737cb8436e3_l3.png" height="26" width="295" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[w(p,f)= U(p,f)-L(p,f)\]" title="Rendered by QuickLaTeX.com"/> <ol start="3"> <li>For any bounded function ``$

f $'' on$ [a, b] $and$ pin P[a,b]

$,</li> </ol> <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-02e73fe8e5bff0946058e3cb035a238f_l3.png" height="26" width="448" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[m(b-a)leq L(p,f)leq U(p,f)leq M(b-a)\]" title="Rendered by QuickLaTeX.com"/> Where,$

m $,$ M $are infimum and supremum of ``$ f $'' on$ [a, b] $. <ul> <li>Definition</li> </ul> Refinement of a partition: Let,$ 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} $''. <ul> <li>Definition:</li> </ul> Common Refinement: If$ 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} $. <ul> <li>Properties of refinement of a partition:</li> </ul> If ``$ f $'' is bounded on$ [a, b] $and$ p_{1}, p_{2} in P [a,b] $where ``$ p_{2} $'' is the refinement of ``$ p_{1} $'' Then begin{align*} text{(i).} &; U(p_{1},f)geq U(p_{2},f)\ text{(ii).} &; L(p_{1},f)leq L(p_{2},f) \ text{(iii).} &; w(p_{1},f)geq w(p_{2},f) end{align*} <h1>Definition :(Darboux Lower Integral)</h1> Let ``$ f $'' be a real valued function over$ [a,b] $, then$ sup{lbrace L(p,f) : pin 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) : pin 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] $begin{align*} text{(i).} &; L(p_{1},f)leq U(p_{2},f)\ text{(ii).} &; int_{ubar{a}}^{b}f(x)dx leq int_{a}^{bar{b}}f(x)dx end{align*} Definition: <h1>Riemann Integrability:</h1> A real valued function ``$ 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,; xin[0,1] $Check$ f(x) in mathcal{R}[0,1]

$Solution: Let <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-0928069e46b35843d01f1f3b5ca65438_l3.png" height="23" width="703" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[p_{n}=leftlbrace 0,dfrac{1}{n},dfrac{2}{n},cdots , dfrac{2}{n} , 1 rightrbrace\]" title="Rendered by QuickLaTeX.com"/> 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] $begin{align*} M_{r} & =dfrac{r}{n}, ;;;;;;; text{ for all } r=1,2,ldots,n,\ m_{r} & =dfrac{r-1}{n}, ;; text{ for all } r=1,2,ldots,n. end{align*} begin{align*} text{Then, } L(p_{n},f) &=sum_{r=1}^{n}m_{r}delta_{r}\ & =sum_{r=1}^{n}dfrac{r-1}{n}timesdfrac{1}{n}\ & =dfrac{1}{n^{2}}sum_{r=1}^{n}(r-1)\ & =dfrac{1}{n^{2}}times dfrac{(n-1)n}{2}\ L(p_{n},f) &=dfrac{1}{2}-dfrac{1}{2n}\ L(p_{n},f) &<dfrac{1}{2} end{align*} Also begin{align*} U(p_{n},f) &=sum_{r=1}^{n}M_{r}delta_{r}\ & =sum_{r=1}^{n}dfrac{r}{n}timesdfrac{1}{n}\ & =dfrac{1}{n^{2}}sum_{r=1}^{n}r\ & =dfrac{1}{n^{2}}times dfrac{(n+1)n}{2}\ U(p_{n},f) &=dfrac{1}{2}+dfrac{1}{2n}\ L(p_{n},f) &>dfrac{1}{2} end{align*} So, [sup{lbrace L(p_{n},f): nin mathbb{N}rbrace}=dfrac{1}{2}] [ int_{ubar{0}}^{1}f(x)dx=dfrac{1}{2} ] And [inf{lbrace U(p_{n},f): nin mathbb{N}rbrace}=dfrac{1}{2}] [ int_{0}^{bar{1}}f(x)dx=dfrac{1}{2} ] Here, [int_{ubar{0}}^{1}f(x)dx=int_{0}^{bar{1}}f(x)dx=dfrac{1}{2}] Hence, [f(x) in mathcal{R}[0,1]] And [int_{0}^{1}f(x)dx=dfrac{1}{2}] So, we understand by the example how we can use the definition of Riemann Integration. But for more complicated functions there will be difficulties to find maximum and minimum in every subinterval so, now we are going to modify the theory of ``when a function is said to be a Riemann integral.'' Theorem : let$ 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] $such that [L(p,f)>intlimits_{ubar{a}}^{b}f(x)dx-dfrac{epsilon}{2}=intlimits_{a}^{b}f(x)dx-dfrac{epsilon}{2}] Also [U(p,f)<intlimits_{a}^{bar{b}}f(x)dx+dfrac{epsilon}{2}=intlimits_{a}^{b}f(x)dx+dfrac{epsilon}{2}] Therefore [U(p,f)-L(p,f)<intlimits_{a}^{b}f(x)dx+dfrac{epsilon}{2}-intlimits_{a}^{b}f(x)dx+dfrac{epsilon}{2}=epsilon] Hence [U(p,f)-L(p,f)<epsilon] Conversely (Sufficient condition) Suppose for any positive number$ epsilon $there exists a partition$ P $on$ [a,b] $such that [U(p,f)-L(p,f)<epsilon] Since, [intlimits_{a}^{bar{b}}f(x)dxleq U(p,f)] and [intlimits_{ubar{a}}^{b}f(x)dxgeq L(p,f)] Also [intlimits_{a}^{bar{b}}f(x)dxgeq intlimits_{ubar{a}}^{b}f(x)dx] Therefore [0leq intlimits_{a}^{bar{b}}f(x)dx-intlimits_{ubar{a}}^{b}f(x)dxleq U(p,f)-L(p,f) Hence Therefore [intlimits_{a}^{bar{b}}f(x)dx=intlimits_{ubar{a}}^{b}f(x)dx] Hence$ 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} $converges to zero and [lim_{ntoinfty}lbrace U(lbrace p_{n}rbrace, f) - L(lbrace p_{n}rbrace, f)rbrace=0] Then$ 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 <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-00c67ee0942dfa775948f0e59df18fa0_l3.png" height="30" width="771" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=begin{cases} x^{2}, ; x text{ x is rational }\ 1, ; x text{ is irrational} end{cases}\]" title="Rendered by QuickLaTeX.com"/> 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] $Let for$ nin mathbb{N}

$we choose a partition <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-0928069e46b35843d01f1f3b5ca65438_l3.png" height="23" width="703" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[p_{n}=leftlbrace 0,dfrac{1}{n},dfrac{2}{n},cdots , dfrac{2}{n} , 1 rightrbrace\]" title="Rendered by QuickLaTeX.com"/> s [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{leftlbrace f(x): xin left[dfrac{r-1}{n},dfrac{r}{n} right] rightrbrace} text{ for all } r=1,2,ldots,n.] [m_{r}=inf{leftlbrace f(x): xin left[dfrac{r-1}{n},dfrac{r}{n} right] rightrbrace} 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, <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-4b29ca7af28916430e63c7fd7d48e976_l3.png" height="23" width="339" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[M_{r}=1 text{ for } r=1,2,ldots,n\]" title="Rendered by QuickLaTeX.com"/> <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-7c1d27d87443146568a25439c03bc3f5_l3.png" height="26" width="446" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[m_{r}=(r-1/n)2 text{ for } r=1,2,ldots,n.\]" title="Rendered by QuickLaTeX.com"/> 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 <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-20a4da3b99020baf60a515cf1ee2878d_l3.png" height="27" width="805" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[lim_{ntoinfty}lbrace U(p_{n},f)-L(p_{n},f) rbrace=1-dfrac{1}{3}=dfrac{2}{3}neq 0.\]" title="Rendered by QuickLaTeX.com"/> Hence by the Remark we can say that$

f $is not Riemann integral on$ [0,1] $. <h1>Properties of Riemann integrable functions : </h1> <ul> <li>If$ f , : , [a,b]to mathbb{R} $is monotonic then$ f in mathbb{R} [a,b] $. i.e$ f $is Riemann integrable.</li> <li>If$ f , : , [a,b]to mathbb{R} $is continuous then$ f $is Riemann integrable.</li> <li>If$ 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] $</li> <li>If$ 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] $.</li> <li>If$ f , : , [a,b]to mathbb{R} $is integrable on$ [a,b] $, then$ f $is integrable on every closed sub interval of$ [a,b] $.</li> <li>If$ a<c </ul> [int_{a}^{c}f(x)dx+int_{c}^{b}f(x)dx=int_{a}^{b}f(x)dx] <ul> <li>If$ f , : , [a,b]to mathbb{R} $is integrable on$ [a,b] $then for every$ lambdainmathbb{R} $then$ lambda f(x) $is also integrable on$ [a,b] $and further more,</li> </ul> [int_{a}^{b}lambda f(x)dx=lambdaint_{a}^{b}f(x)dx] <ul> <li>If$ f; g , : , [a,b]to mathbb{R} $two integrable on$ [a,b] $and</li> <li>If$ 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] $furthermore,</li> </ul> [int_{a}^{b}lbrace lambda f(x)+mu g(x)rbrace dx=lambdaint_{a}^{b}f(x)dx+muint_{a}^{b}g(x)dx] (x) If$ 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?) <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-beb6a596eabcfad5b26bc87c53df092d_l3.png" height="26" width="959" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[text{Consider }f(x)=begin{cases} ;;; 1, ; x text{ x is rational }\ -1, ; x text{ is irrational} end{cases}\]" title="Rendered by QuickLaTeX.com"/> 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] $. <ul> <li>If$ 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] $.</li> <li>If$ 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</li> </ul> <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-af991928da37d54f258329d347b013d1_l3.png" height="26" width="823" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=begin{cases} ;;; 1, ; x text{ x is rational }\ -1, ; x text{ is irrational} end{cases}\]" title="Rendered by QuickLaTeX.com"/> <ul> <li>If$

f , : , [a,b]to mathbb{R} $and$ g , : , [a,b]to mathbb{R} $integrable and$ g(x) geq k ; forall xin [a,b] $where$ k>0 $, then$ f/g $also integrable as$ [a,b] $.</li> <li>If$ f , : , [a,b]to mathbb{R} $is Riemann integrable on$ [a,b] $and$ f(x) geq k ; forall xin [a,b] $where$ k>0 $then$ 1/f $also integrable on$ [a,b] $.</li> <li>If$ I=[a,b] $and$ J=[c,d] $bounded and closed interval. Let$ f , : , Ito mathbb{R} $and$ f , : , Jto mathbb{R} $integrable and continuous function such that$ f(I) subset [c,d] $then$ gcirc f : Ito mathbb{R} $is integrable on$ I $.</li> </ul> Example 1 : - let$ f:[0,1]to mathbb{R} $be a bounded function on$ [0,1]

$defined by <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-c67b6a9199e71e7883f7e2236aa82908_l3.png" height="83" width="1841" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=begin{cases} 1, ;;;;;;;; x =0\ x, ; 0 Now we are going to check $ dfrac{1}{f(x)} $ is Riemann Integrable or not. Let's start:- Here $f:[0,1]to mathbb{R}$ is bounded function also $f$ is continuous on $[0,1]$ except $x=0$, so $f$ is Riemann Integrable on $[0,1]$. But\]" title="Rendered by QuickLaTeX.com"/>dfrac{1}{f(x)}=begin{cases} 1, ;;;;;;;; x =0\ dfrac{1}{x}, ; 0As$

xto 0+ $,$ dfrac{1}{f(x)}toinfty + $, so$ 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 <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-34c5f6ae35cb656a3db63c8d40c82166_l3.png" height="26" width="154" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=x-[x]\]" title="Rendered by QuickLaTeX.com"/> 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 <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-8fc27a3fa559e3e5b828027b977ba8fd_l3.png" height="26" width="1063" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=begin{cases} x,& ;; 0leq x< 1\ x-1,& ; 1leq x<2\ x-2,& 2leq x <3\ x-3,& 3leq x <4\ 0,& x=4 end{cases}\]" title="Rendered by QuickLaTeX.com"/> Since$

0leq 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] $. Let we define begin{align*} g_{1}(x) &=x text{ on } 0leq x <1\ g_{2}(x) &=x-1 text{ on } 1leq x <2\ g_{3}(x) &=x-2 text{ on } 2leq x <3\ g_{4}(x) &=x-3 text{ on } 3leq x <4\ end{align*} And then$ f(x)

$becomes <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-4baf3137848ca9b89e06ae8460b93827_l3.png" height="26" width="1137" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=begin{cases} g_{1}(x),& ;; 0leq x< 1\ g_{2}(x),& ; 1leq x<2\ g_{3}(x),& 2leq x <3\ g_{4}(x),& 3leq x <4\ 0,& x=4 end{cases}\]" title="Rendered by QuickLaTeX.com"/> 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) $is Riemann Integrable and find$ displaystyleint_{0}^{4}f(x)dx $(Answer$ displaystyleint_{0}^{4}f(x)dx=17 $) <h1>Some Inequalities related to Riemann Integration:</h1> <ol> <li>If$ f:[a,b]to mathbb{R} $is integrable on$ [a, b] $and$ f(x)geq 0 $for all values of$ x $in$ [0, 1]

$.</li> </ol> Then <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-81890e6ce439fdaf0c337a7fe69de2e6_l3.png" height="31" width="171" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[int_{a}^{b}f(x)dxgeq 0\]" title="Rendered by QuickLaTeX.com"/> <ol start="2"> <li>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] $.</li> </ol> Then [int_{a}^{b}f(x)dxgeq int_{a}^{b}g(x)dx] <ol start="3"> <li>If$ f : [a,b]to mathbb{R} $is integrable on$ [a, b] $</li> </ol> Then [leftvertint_{a}^{b}f(x)dxrightvertleq int_{a}^{b}vert f(x)vert dx] Problems related with these theories:- Example. Show that [dfrac{1}{2}leqint_{0}^{1}dfrac{dx}{sqrt{4-x^{2}+x^{3}}}leq dfrac{pi}{6} ] To prove this we are going to use all of the theories which we learnt till now, Let [g(x)=dfrac{1}{sqrt{4-x^{2}+x^{3}}}, xin[0,1] ] Here for all$ xin[0,1]

$ <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-97cc1c0d0602688869b438ec14ed2c67_l3.png" height="29" width="170" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[0leq x^{3}leq x^{2} leq 1\]" title="Rendered by QuickLaTeX.com"/> For all$

xin[0,1]

$ <img src="https://www.mathacademytutoring.com/wp-content/ql-cache/quicklatex.com-5176a402abe7c8f86566b94970f87172_l3.png" height="29" width="322" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[0leq 4-x^{2}leq 4- x^{2}+ x^{3}leq 4,\]" title="Rendered by QuickLaTeX.com"/> For all$

xin[0,1] $[dfrac{1}{2}leqdfrac{1}{sqrt{4-x^{2}+x^{3}}}leq dfrac{1}{sqrt{4-x^{2}}}] Therefore let for all$ xin[0,1] $[f(x)=dfrac{1}{sqrt{4-x^{2}}}] and [h(x)=dfrac{1}{2}] Here for all$ xin[0,1] $[f(x)geq g(x) geq h(x)] So therefore [int_{0}^{1}f(x)dxgeq int_{0}^{1}g(x)dxgeq int_{0}^{1}h(x)dx] [int_{0}^{1}dfrac{1}{sqrt{4-x^{2}}}dxgeq int_{0}^{1}dfrac{1}{sqrt{4-x^{2}+x^{3}}}dxgeq int_{0}^{1}dfrac{1}{2}dx] [int_{0}^{1}dfrac{1}{2}dxleq int_{0}^{1}dfrac{1}{sqrt{4-x^{2}+x^{3}}}dxleqint_{0}^{1}dfrac{1}{sqrt{4-x^{2}}}dx] Now [int_{0}^{1}dfrac{1}{sqrt{4-x^{2}}}dx=left[sin^{-1}left(dfrac{x}{2}right)right]_{0}^{1}] ` [int_{0}^{1}dfrac{1}{sqrt{4-x^{2}}}dx=dfrac{pi}{6}] And [int_{0}^{1}dfrac{1}{2}dx=dfrac{1}{2}] Therefore [int_{0}^{1}dfrac{1}{2}dxleq int_{0}^{1}dfrac{1}{sqrt{4-x^{2}+x^{3}}}dxleqint_{0}^{1}dfrac{1}{sqrt{4-x^{2}}}dx] [dfrac{1}{2}leq int_{0}^{1}dfrac{1}{sqrt{4-x^{2}+x^{3}}}dxleqdfrac{pi}{6}] Hence proved. Now some work sheet problems:- <ol> <li>Let [f(x)=x^{2}, xin[0,1]]</li> </ol> Check$ f(x)$ is Riemann Integrable or not.

Try to prove these problems

i) [dfrac{1}{3sqrt{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}}dxleq 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}xdxleq 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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