# Riemann Integral

by | Nov 26, 2021 | Math Learning

# 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 ().

• Definition:

Partition: Let be a given closed and bounded on real line. Let be finite number of points in such that P = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbrace[a, b][a, b]n[ x_{i-1}, x_{i} ]i=1,2,ldots, nP[a, b]ab[a, b]P = [a, b][a, b][a, b][a, b][a, b]P [a, b]pin P[a, b]p[a, b]np_{1}=leftlbrace 0,dfrac{1}{n},dfrac{2}{n},cdots , dfrac{n-1}{n} , 1 rightrbrace[0,1]p_{2}=leftlbrace 0,dfrac{1}{2n},dfrac{2}{2n},cdots , dfrac{2n-1}{2n} , 1 rightrbrace[0,1]p_{3}=leftlbrace 0,dfrac{1}{3n},dfrac{2}{3n},cdots , dfrac{3n-1}{3n} , 1 rightrbrace[0,1]1notin PP = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbrace[a, b]I_{r}= [x_{r-1},ldots, x_{r},][a, b]r=1, 2, …, n delta r = vert I_{r}vert =(x_{r}-x_{r-1}) r=1, 2, ldots, nmax lbrace vert I_{r} vert ; : ; r=1, 2, ldots, n rbrace pdeltaVert P Vertdelta = max lbrace delta_{r} : r = 1, 2, ldots, nrbrace f:[a,b]tomathbb{R}[a, b]P = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbrace[a, b]f[ x_{r-1}, x_{r} ] (r=1, 2, ldots, n)M_{r}=suplbrace f(x) ; : ; x in [ x_{r-1}, x_{r} ] rbracedisplaystylesum_{r=1}^{n}M_{r}delta_{r}f:[a,b]tomathbb{R}[a, b]P = lbrace x_{0}, x_{1}, x_{2}, ldots, x_{n} rbrace[a, b]m_{r}=inflbrace f(x) ; : ; x in [ x_{r-1}, x_{r} ] rbracedisplaystylesum_{r=1}^{n}m_{r}delta_{r}fp[a, b]

fp[a, b]

U(p,f)L(p,f)f(x)

fp[a, b]

f[a, b]pin P[a,b]

mMf[a, b]p_{1}p_{2}[a, b]p_{1}subset p_{2} text{ and } delta(p_{1})leq delta(p_{2})p_{2}p_{1}p_{1}p_{2}[a, b]p_{1}subset p_{1}cup p_{2}p_{2}subset p_{1}cup p_{2}p_{1}cup p_{2}p_{1}p_{2}p_{1}p_{2}f[a, b]p_{1}, p_{2} in P [a,b]p_{2}p_{1}f[a,b]sup{lbrace L(p,f)  :  pin P [a,b]rbrace}f[a,b]f[a, b]inf{lbrace U(p,f) :  pin P  [a,b] rbrace}f[a, b][a, b]p_{1}p_{2}[a, b]f[a, b][a, b]ff f in mathcal{R}[a,b] f(x)=x,; xin[0,1] f(x) in mathcal{R}[0,1]

[0, 1]np_{n}[0, 1]nf[0, 1]f: [a,b] to mathbb{R}[a,b][a,b]epsilon > 0 , exists text{ partition } P text{ of } [a,b] f[a,b] epsilon f[a,b] epsilon P[a,b]epsilonP[a,b]f(x)[a,b]f:[a,b] to mathbb{R}[a,b]{p_{n}}[a,b]{vert p_{n}vert}f[a,b]f:[0,1] to mathbb{R}

f(x) text{ on } [0,1]f(x)[0,1]nin mathbb{N}

left[dfrac{r-1}{n},dfrac{r}{n} right]r=1,2,ldots,n.

f[0,1]f , : , [a,b]to mathbb{R}f in mathbb{R} [a,b]ff , : , [a,b]to mathbb{R}ff , : , [a,b]to mathbb{R}[a,b]f in mathbb{R} [a,b]f , : , [a,b]to mathbb{R}[a,b][a,b]f(x)in mathcal{R}[0,1]f , : , [a,b]to mathbb{R}[a,b]f[a,b]a<c<bf , : , [a,b]to mathbb{R}[a,c][a,b]f , : , [a,b]to mathbb{R}[a,b] lambdainmathbb{R} lambda f(x)[a,b]f; g , : , [a,b]to mathbb{R}[a,b]f , : , [a,b]to mathbb{R}g , : , [a,b]to mathbb{R} lambda,mu in mathbb{R} ; lambda f(x)+mu g(x) [a,b]f , : , [a,b]to mathbb{R}[a,b]vert f(x) vert[a,b]

f(x)vert f(x)vert= 1;; forall x in [a,b]vert f(x)vert[a,b]f , : , [a,b]to mathbb{R}g , : , [a,b]to mathbb{R}[a,b]fg[a,b]f , : , [a,b]to mathbb{R}[a,b]f^{2}[a,b]

f , : , [a,b]to mathbb{R}g , : , [a,b]to mathbb{R} g(x) geq k ; forall xin [a,b] k>0 f/g[a,b]f , : , [a,b]to mathbb{R}[a,b] f(x) geq k ; forall xin [a,b] k>0 1/f [a,b]I=[a,b]J=[c,d]f , : , Ito mathbb{R}f , : , Jto mathbb{R}f(I) subset [c,d] gcirc f : Ito mathbb{R} If:[0,1]to mathbb{R}[0,1]

xto 0+ dfrac{1}{f(x)}toinfty +dfrac{1}{f(x)}[0,1]dfrac{1}{f(x)}[0,1]f(x):[0,4]to mathbb{R}

f(x)[0,4]f(x)

0leq f(x)<1x[0,4]f(x)[0,4]f(x)[0,4]x=1, x=2, x=3,x=4f(x)[0,4]f[0,4]f(x)[0,4]f(x)[0,4]f(x)

f(x):[0,4]to mathbb{R}f(x) displaystyleint_{0}^{4}f(x)dx displaystyleint_{0}^{4}f(x)dx=17 f:[a,b]to mathbb{R}[a, b]f(x)geq 0x[0, 1]

f:[a,b]to mathbb{R}g : [a,b]to mathbb{R}[a, b]f(x)geq g(x)x[0, 1]f : [a,b]to mathbb{R}[a, b]xin[0,1]

xin[0,1]

xin[0,1]xin[0,1]xin[0,1]f(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 ]

# Bibliography

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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