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Riemann Integral

by | 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}P = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbraceis 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.   <u>Fact</u>: 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><u>Definition: </u></li> </ul> <u>Family of partition</u>: 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 byP [a, b].Thus, ifpin P[a, b]implies ``p'' is a partition of[a, b]. <u>Example:</u> For positive integer ``n'',p_{1}=leftlbrace 0,dfrac{1}{n},dfrac{2}{n},cdots , dfrac{n-1}{n} , 1 rightrbraceis a partition of[0,1]p_{2}=leftlbrace 0,dfrac{1}{2n},dfrac{2}{2n},cdots , dfrac{2n-1}{2n} , 1 rightrbraceis 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><u>Definition</u>:</li> </ul> <u>Norm of partition</u>: IfP = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbracebe a partition of[a, b]andI_{r}= [x_{r-1},ldots, x_{r},]denotes the sub-intervals of[a, b](forr=1, 2, …, n) having the length delta r = vert I_{r}vert =(x_{r}-x_{r-1}) ,  forr=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 bydeltaorVert P Vert. i,e.,  delta = max lbrace delta_{r} : r = 1, 2, ldots, nrbrace . <ul>  	<li><u>Definition</u>:</li> </ul> <u>Upper Sum</u>: Let,f:[a,b]tomathbb{R}be a bounded function on[a, b]andP = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbracebe 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} ] rbraceThen,displaystylesum_{r=1}^{n}M_{r}delta_{r}defined as upper sum or Darboux upper sum.<sub> r</sub> <u>Lower sum:</u> Let,f:[a,b]tomathbb{R}be a bounded function on[a, b]andP = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbracebe 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. <sub>r</sub> <u>Notation:</u> <u>Upper Sum</u>: For the function ``f'' over the partition''p'' on[a, b]

    , Then, <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> <u>Lower Sum</u>: For the function ``

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

    , Then, <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> <u> Facts:</u> <ol>  	<li>If

U(p,f)andL(p,f)are equal whenf(x)

    is a constant function.</li> </ol> 2. <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> is the 'oscillatory sum' of ``

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

    , and we write that as <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> <ol start="3">  	<li>For any bounded function ``

f'' on[a, b]andpin P[a,b]

    ,</li> </ol> <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> Where,

m,Mare infimum and supremum of ``f'' on[a, b]. <ul>  	<li><u>Definition</u></li> </ul> <u>Refinement of a partition</u>: Let,p_{1}andp_{2}be two partitions of[a, b]such that ifp_{1}subset p_{2} text{ and } delta(p_{1})leq delta(p_{2})thenp_{2}is called refinement of the partition ``p_{1}''. <ul>  	<li><u>Definition</u>:</li> </ul> <u>Common Refinement:</u> Ifp_{1}andp_{2}are two partitions of[a, b]thenp_{1}subset p_{1}cup p_{2}andp_{2}subset p_{1}cup p_{2}. Sop_{1}cup p_{2}is the refinement of bothp_{1}andp_{2}and is called common refinement ofp_{1}andp_{2}. <ul>  	<li><u>Properties of refinement of a partition:</u></li> </ul> If ``f'' is bounded on[a, b]andp_{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><u>Definition</u> :(<u>Darboux Lower Integral</u>)</h1> Let ``f'' be a real valued function over[a,b], thensup{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  ] <u>Darboux Upper Integral</u>: For real valued function ``f'' on[a, b], theninf{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  ] <u>Particular fact</u>: For any real valued bounded function on[a, b]and any two partitionsp_{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*} <u>Definition</u>: <h1><u>Riemann Integrability</u>:</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] <u>Example</u>: Let f(x)=x,; xin[0,1] Check f(x) in mathcal{R}[0,1]

    <u>Solution</u>:   Let <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> 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.] Herefis 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.'' <u>Theorem :</u> letf: [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 positionepsilon > 0 , exists text{ partition } P text{ of } [a,b] such that [U(p,f)-L(p,f)<epsilon] <u>Proof : </u> (Necessary Part:) Letfbe 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 offon[a,b], we have for the given positive epsilon , there exist a partitionPon[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 numberepsilonthere exists a partitionPon[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)] <u>A</u>lso [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] Hencef(x)is Riemann Integrable on[a,b]Hence proved. <u>Remark:-</u> letf:[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] Thenfis Riemann integrable on[a,b]. Now we are going to learn how to use the Remark Example : letf:[0,1] to mathbb{R}

    be a function by <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> Here we are going to check the integrability of

f(x) text{ on } [0,1]<u>Lets try:</u> Heref(x)is bounded on[0,1]Let fornin mathbb{N}

    we choose a partition <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> 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 eachr=1,2,ldots,n.

    So, <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> 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 <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> Hence by the Remark we can say that

fis not Riemann integral on[0,1]. <h1><u>Properties of Riemann integrable functions : </u></h1> <ul>  	<li>Iff , : , [a,b]to mathbb{R}is monotonic thenf in mathbb{R} [a,b]. i.efis Riemann integrable.</li>  	<li>Iff , : , [a,b]to mathbb{R}is continuous thenfis Riemann integrable.</li>  	<li>Iff , : , [a,b]to mathbb{R}be bounded but has finite number of discontinuous points on[a,b]thenf in mathbb{R} [a,b]</li>  	<li>Iff , : , [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]thenf(x)in mathcal{R}[0,1].</li>  	<li>Iff , : , [a,b]to mathbb{R} is integrable on[a,b], thenfis integrable on every closed sub interval of[a,b].</li>  	<li>Ifa<c<b, andf , : , [a,b]to mathbb{R}is integrable on[a,c]and on[a,b]then</li> </ul> [int_{a}^{c}f(x)dx+int_{c}^{b}f(x)dx=int_{a}^{b}f(x)dx] <ul>  	<li>Iff , : , [a,b]to mathbb{R}is integrable on[a,b]then for every lambdainmathbb{R} thenlambda 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>Iff; g , : , [a,b]to mathbb{R}two integrable on[a,b]and</li>  	<li>Iff , : , [a,b]to mathbb{R}andg , : , [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)           Iff , : , [a,b]to mathbb{R} is integrable on[a,b], thenvert f(x) verton[a,b]

    integrable. But the converse is not true. (Why?) <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> Here,

f(x)is not Riemann integrable butvert f(x)vert= 1;; forall x in [a,b]vert f(x)vertis constant then it must be Riemann integrable on[a,b]. <ul>  	<li>Iff , : , [a,b]to mathbb{R}andg , : , [a,b]to mathbb{R}are both integrable on[a,b], thenfgalso integrable on[a,b].</li>  	<li>Iff , : , [a,b]to mathbb{R}is integrable on[a,b], thesef^{2}is integrable on[a,b]

    but the converse is not true (try with the  same function</li> </ul> <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> <ul>  	<li>If

f , : , [a,b]to mathbb{R}andg , : , [a,b]to mathbb{R}integrable and g(x) geq k ; forall xin [a,b] wherek>0 , then f/g  also integrable as[a,b].</li>  	<li>Iff , : , [a,b]to mathbb{R}is Riemann integrable on[a,b]and f(x) geq k ; forall xin [a,b] wherek>0 then1/f also integrable on[a,b].</li>  	<li>IfI=[a,b]andJ=[c,d]bounded and closed interval. Letf , : , Ito mathbb{R}  andf , : , Jto mathbb{R} integrable and continuous function such thatf(I) subset [c,d]then gcirc f : Ito mathbb{R}    is integrable onI.</li> </ul> Example 1 : -  letf:[0,1]to mathbb{R} be a bounded function on[0,1]

    defined by <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/>dfrac{1}{f(x)}=begin{cases} 1, ;;;;;;;; x =0\ dfrac{1}{x}, ; 0As

xto 0+ ,dfrac{1}{f(x)}toinfty +, sodfrac{1}{f(x)}is not bounded on[0,1]and hencedfrac{1}{f(x)}is not Riemann Integrable on[0,1]  and we are done. Example 2 Consider the functionf(x):[0,4]to mathbb{R}

    such that <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" height="26" width="154" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[f(x)=x-[x]\]" title="Rendered by"/> 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 off(x)

    is <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> Since

0leq f(x)<1for allxin[0,4],f(x)is bounded on[0,4].f(x)is continuous on[0,4]except the pointsx=1, x=2, x=3,andx=4. Sof(x)has finite number of discontinuity on[0,4]then we can say thatfis integrable on[0,4]So hencef(x)is Riemann Integrable on[0,4] Now we are going to find the value of the integration of the functionf(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  thenf(x)

    becomes <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> 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 provef(x)is Riemann Integrable and find displaystyleint_{0}^{4}f(x)dx   (Answer displaystyleint_{0}^{4}f(x)dx=17 ) <h1><u>Some Inequalities related to Riemann Integration:</u></h1> <ol>  	<li>Iff:[a,b]to mathbb{R}is integrable on[a, b]andf(x)geq 0for all values ofxin[0, 1]

    .</li> </ol> Then <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" height="31" width="171" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[int_{a}^{b}f(x)dxgeq 0\]" title="Rendered by"/> <ol start="2">  	<li>If

f:[a,b]to mathbb{R}andg : [a,b]to mathbb{R}is integrable on[a, b]andf(x)geq g(x)for all values ofxin[0, 1].</li> </ol> Then [int_{a}^{b}f(x)dxgeq int_{a}^{b}g(x)dx] <ol start="3">  	<li>Iff : [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] <u>Problems related with these theories:-</u> <u> </u><u>Example.</u>                         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 allxin[0,1]

    <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" height="29" width="170" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[0leq x^{3}leq x^{2} leq 1\]" title="Rendered by"/>  For all


     <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="" 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"/> 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 allxin[0,1][f(x)=dfrac{1}{sqrt{4-x^{2}}}] and [h(x)=dfrac{1}{2}] Here for allxin[0,1]<u>[f(x)geq g(x) geq h(x)]</u> 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> Checkf(x)$ is Riemann Integrable or not.

  1. 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}]

  1. ii)

[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}]

  1. Try to show that

[int_{0}^{frac{pi}{2}}sin^{n+1}xdxleq int_{0}^{frac{pi}{2}}sin^{n} xdx ]  


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

  Thank you………

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