## What is Riemann Integral?

In calculus, as the partitions get finer, the Riemann integral is the limit of the Riemann sums of a function. A function is said to be integrable (or Riemann-integrable) if the limit of the Riemann sums is defined. If a function g is Riemann integrable in interval $\left[c,d\right]$, then g is said to be belonging to $R\left[a,b\right]$. By making the partition fine enough, the Riemann sum can be rendered as similar to the Riemann integral as needed. It is one of the most important aspects of calculus.

## Riemann Integral Explained

Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.

Let's pretend that g is a function that is defined over the closed interval $\left[c,d\right]$. Assume that g is a continuous, non-negative function. The region between the graph of g and the x-axis is defined by the integral of g with respect to x. This region is known as the definite integral of the function g from point c to point d.

In calculus, consider g as a non-negative real-valued function defined on interval $\left[c,d\right]$ and let $S=\left\{\left(x,y\right):c\le x be the plane region above the interval [a, b] and under the graph of the function g. The area of S is something we'd like to know about. We'll denote the area by integral$\underset{a}{\overset{b}{\int }}g\left(x\right)dx$ after we've measured it.

The Riemann integral is based on the use of very simple approximations for the region S.

We may claim that "in the limit," we get exactly the region S lying under the curve by taking better and better approximations.

When g can be both positive and negative, the integral corresponds to the signed area under the graph of g, that is the area below the x-axis subtracted from the area above the x-axis.

### Glossary for Riemann Integral

• Partition: A set of points $P=\left\{c={t}_{0}<{t}_{1}<{t}_{2}<\cdots {t}_{k-1}<{t}_{k}<\cdots {t}_{n}=d\right\}$ is called a partition of $\left[c,d\right]$.
• Length of subinterval: For an rth sub interval $\left[{x}_{r-1},{x}_{r}\right]$ the length of the subinterval is denoted as $\Delta r={S}_{r}={x}_{r}-{x}_{r-1}$
• Norm (mesh) of  a partition (P):

Mesh$P=\mathrm{max}\left\{{S}_{r}:r=0,1,2,\dots ,n\right\}$ where ${S}_{r}={x}_{r}-{x}_{r-1}$.

Note: ${S}_{r}\le$ mesh P

• Refinement of a partition: Let P and Q be any two partitions of interval $\left[c,d\right]$ such that $P\subset Q$ then P is called a refinement of Q.

## Riemann Sum

A Riemann sum is a calculation that approximates the area of a region by adding the areas of several condensed slices of the region. It is used in calculus to formalize the exhaustion equation, which is used to calculate the area of a field. This procedure yields the integral, which precisely computes the area value.

Mathematically, the Riemann sum is defined as an approximation, in which we simply sum up the areas of all the rectangles to estimate the region under the graph of f. The sum of the areas of all n rectangles is written in summation notation for $i=0,1,\dots ,n-1$ as:

$\sum _{i=0}^{n-1}f\left({x}_{i}\right)\Delta {x}_{i}$

This is called Riemann sum.

To make the Riemann sum approximation better we increase the number of subintervals and in order to do so, we decrease the width of each subinterval.

The Riemann sum is of two types, the left Riemann sum and the right Riemann sum.

The left Riemann sum states that the function should be evaluated at the subinterval's left-hand endpoint and the rectangle should be of that height.

The right Riemann sum states that the function should be evaluated at the subinterval's right-hand endpoint and the rectangle should be of that height.

### Definite Integral in Terms of Riemann Sum

While computing Riemann sum, when the number of intervals tends to infinity, that is if we make a large number of intervals, we partition the interval into tiny subintervals then we can express Riemann sum as a definite integral as:

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{n\to \infty }{\mathrm{lim}}\sum _{i=0}^{n-1}f\left({x}_{i}\right)\Delta {x}_{i}$

It is not necessary to compute the Riemann sum by diving the area under the curve using rectangles we can also make trapezoids to compute the Riemann sum and thus the integral as well.

One of the applications of the Riemann sum is to evaluate the pressure exerted on a submerged object.

### Properties of Riemann Integral

1) Every monotonic function g on $\left[c,d\right]$ is Riemann integrable.

Proof: Let g be a monotonic function in any interval $\left[c,d\right]$. We have to prove that g satisfies the condition of Riemann integrability.

Since g is monotonic in $\left[c,d\right]$, therefore, either it is monotonic increasing or monotonic decreasing in $\left[c,d\right]$.

Case 1: Let g is monotonic increasing in $\left[c,d\right]$, that is $g\left(c\right)0$.

Now we select a partition P of $\left[c,d\right]$ such that:

$P=\left\{c={t}_{0}<{t}_{1}<{t}_{2}<\cdots {t}_{k-1}<{t}_{k}<\cdots {t}_{n}=d\right\}$

With mesh $P<\frac{\epsilon }{g\left(d\right)-g\left(c\right)}$ ……(1)

It is required to prove that $U\left(g,P\right)-L\left(g,P\right)<\epsilon$.

Now, we have,

U g,P L g,P = k=1 n M k δ k k=1 n m k δ k = k=1 n M g t k1 , t k t k t k1 k=1 n m g t k1 , t k t k t k1 = k=1 n M g t k1 , t k m g t k1 , t k t k t k1

Using equation (1) we can rewrite the above expression as,

$U\left(g,P\right)-L\left(g,P\right)<\frac{\epsilon }{g\left(d\right)-g\left(c\right)}\sum _{k=1}^{n}\left\{M\left[g\left[{t}_{k-1},{t}_{k}\right]\right]-m\left[g\left[{t}_{k-1},{t}_{k}\right]\right]\right\}\left({t}_{k}-{t}_{k-1}\right)$

Since function g is increasing, we have,

$\begin{array}{l}{M}_{k}=\mathrm{sup}\left\{g,\left[{t}_{k-1},{t}_{k}\right]\right\}=g\left({t}_{k}\right)\\ {m}_{k}=\mathrm{inf}\left\{g,\left[{t}_{k-1},{t}_{k}\right]\right\}=g\left({t}_{k-1}\right)\end{array}$

On further simplification we obtain,

U g,P L g,P = ε g d g c k=1 n g t k g t k1 = ε g d g c g d g c =ε

Thus, the relation obtained is $U\left(g,P\right)-L\left(g,P\right)<\epsilon$.

Hence, the function g satisfies the condition of Riemann integrability.

Case 2: Let g is monotonic decreasing in $\left[c,d\right]$ that is $g\left(c\right)>g\left(d\right)⇒g\left(c\right)-g\left(d\right)>0$.

Now we select a partition P of $\left[c,d\right]$ such that:

$P=\left\{c={t}_{0}>{t}_{1}>{t}_{2}>\cdots {t}_{k-1}>{t}_{k}>\cdots {t}_{n}=d\right\}$

With mesh $P<\frac{\epsilon }{g\left(c\right)-g\left(d\right)}$ ……(2)

It is required to prove that $U\left(g,P\right)-L\left(g,P\right)<\epsilon$.

Now, we have,

U g,P L g,P = k=1 n M k δ k k=1 n m k δ k = k=1 n M g t k1 , t k t k t k1 k=1 n m g t k1 , t k t k t k1 = k=1 n M g t k1 , t k m g t k1 , t k t k t k1

Using equation (2) we can rewrite the above expression as,

$U\left(g,P\right)-L\left(g,P\right)<\frac{\epsilon }{g\left(c\right)-g\left(d\right)}\sum _{k=1}^{n}\left\{M\left[g\left[{t}_{k-1},{t}_{k}\right]\right]-m\left[g\left[{t}_{k-1},{t}_{k}\right]\right]\right\}\left({t}_{k}-{t}_{k-1}\right)$

Since function g is decreasing, we have,

$\begin{array}{l}{M}_{k}=\mathrm{sup}\left\{g,\left[{t}_{k-1},{t}_{k}\right]\right\}=g\left({t}_{k-1}\right)\\ {m}_{k}=\mathrm{inf}\left\{g,\left[{t}_{k-1},{t}_{k}\right]\right\}=g\left({t}_{k}\right)\end{array}$

On further simplification we obtain,

U g,P L g,P = ε g c g d k=1 n g t k1 g t k = ε g c g d g c g d =ε

Thus, the relation obtained is $U\left(g,P\right)-L\left(g,P\right)<\epsilon$.

Hence, the function g satisfies the condition of Riemann integrability.

2) Every continuous function g in $\left[c,d\right]$ is Riemann integrable.

Proof: Let g be a continuous function in a closed interval $\left[c,d\right]$. We have to prove that g satisfies the condition of Riemann integrability.

Since g is continuous in $\left[c,d\right]$, this implies that g is uniformly continuous in $\left[c,d\right]$.

Therefore, for $\epsilon >0$ there exists $\delta >0$ such that for x, y $\in \left[c,d\right]$ such that $\left|x-y\right|<\delta$,$\left|g\left(x\right)-g\left(y\right)\right|<\frac{\epsilon }{d-c}$ ……(3)

Now we select a partition P of $\left[c,d\right]$ such that:

$P=\left\{c={t}_{0}<{t}_{1}<{t}_{2}<\cdots {t}_{k-1}<{t}_{k}<\cdots {t}_{n}=d\right\}$

With mesh $P<\delta$ ……(4)

Since, g is continuous in $\left[c,d\right]$, therefore, it is continuous in each subinterval $\left[{t}_{k-1},{t}_{k}\right]$, for all $k=1,2,\dots ,n$.

Therefore, g assumes its maximum and minimum in each subinterval $\left[{t}_{k-1},{t}_{k}\right]$. That is,

$\left\{M\left[g\left[{t}_{k-1},{t}_{k}\right]\right]-m\left[g\left[{t}_{k-1},{t}_{k}\right]\right]\right\}<\frac{\epsilon }{d-c}$ ……(5)

It is required to prove that $U\left(g,P\right)-L\left(g,P\right)<\epsilon$

We have,

U g,P L g,P = k=1 n M k δ k k=1 n m k δ k = k=1 n M g t k1 , t k t k t k1 k=1 n m g t k1 , t k t k t k1 = k=1 n M g t k1 , t k m g t k1 , t k t k t k1

Simplify it further as,

U g,P L g,P = ε dc k=1 n t t k1 = ε dc dc =ε

Thus, the relation obtained is $U\left(g,P\right)-L\left(g,P\right)<\epsilon$.

Hence, the function g satisfies the condition of Riemann integrability.

### Algebra of Riemann Integral

If f and g both are Riemann integrable in $\left[c,d\right]$ then we have the following:

1) Riemann integrability of a sum: $f+g$ also belongs to $R\left[c,d\right]$. Moreover,

$\underset{c}{\overset{d}{\int }}\left(f+g\right)dx=\underset{c}{\overset{d}{\int }}fdx+\underset{c}{\overset{d}{\int }}gdx$

2) Riemann integrability of subtraction: $f-g$ also belongs to $R\left[c,d\right]$. Moreover,

$\underset{c}{\overset{d}{\int }}\left(f-g\right)dx=\underset{c}{\overset{d}{\int }}fdx-\underset{c}{\overset{d}{\int }}gdx$

3) Riemann integrability of multiplication: $f\cdot g$ and ${f}^{2}$ also belong to $R\left[c,d\right]$.

4) Riemann integrability of division: $\frac{f}{g}$ belongs to $R\left[c,d\right]$, where $g\left(x\right)$ is non zero.

5) Riemann integrability of absolute value: $\left|f\right|$ belongs to $R\left[c,d\right]$. Moreover,

$\left|\underset{c}{\overset{d}{\int }}fdx\right|\le \underset{c}{\overset{d}{\int }}\left|f\right|dx$

6) Riemann integrability of constant function: If f is Riemann integrable, then $kf$is also Riemann integrable, for all k belonging to set of real numbers. Moreover,

$\underset{c}{\overset{d}{\int }}\left(kf\right)dx\le k\underset{c}{\overset{d}{\int }}fdx$

### How is Riemann-Stieltjes Integral Different from Riemann Integral?

The Riemann–Stieltjes integral, named after Bernhard Riemann and Thomas Joannes Stieltjes, is a generalization of the Riemann integral in mathematics. Stieltjes published the first description of this integral in 1894. It is a useful and instructive precursor to the Lebesgue integral, as well as a useful method for unifying analogous types of statistical theorems that refer to discrete arithmetic.

In the original formulation of F. Riesz's theorem, the Riemann–Stieltjes integral is used to describe the dual space of the Banach space C[a,b] of continuous functions in an interval [a,b] as Riemann–Stieltjes integral against functions of bounded variation. That theorem was later rewritten in terms of measurements.

Let g be bounded on $\left[c,d\right]$ and let $P=\left\{c={t}_{0}<{t}_{1}<{t}_{2}<\cdots {t}_{k-1}<{t}_{k}<\cdots {t}_{n}=d\right\}$.

A Riemann-Stieltjes sum of g associated with P and G is a sum of the form:

${J}_{G}\left(g,P\right)+\sum _{k=1}^{n}g\left({x}_{k}\right)\left[G\left({t}_{k}^{-}\right)-G\left({t}_{k-1}^{-}\right)\right]$

Here ${x}_{k}\in \left[{t}_{k-1},{t}_{k}\right]$ for $k=1,2,\dots ,n$.

The function g is Riemann-Stieltjes integrable on $\left[c,d\right]$ if there exists r in set of real numbers with the following property: for each $\epsilon >0$ there exists $\delta >0$ such that $\left|S-r\right|<\epsilon$ for every Riemann-Stieltjes sum S of g associated with a partition P having G-mesh$\left(P\right)<\delta$. We call r the Riemann-Stieltjes integral of g and it is temporarily expressed as integral,

A theorem based on Riemann-Stieltjes integral is: “A bounded function g on $\left[c,d\right]$ is G-integrable if and only if it is Riemann-Stieltjes integrable, in which case the integrals are equal.”

Integration by parts in the form of the Riemann–Stieltjes integral is possible and is expressed as following integral,

### Extension of Riemann Integrability

Lebesgue’s criterion for Riemann integrability: A bounded function $g:\left[c,d\right]\to ℝ$ is Riemann integrable on $\left[c,d\right]$ if and only if the set of points of discontinuity of g in $\left[c,d\right]$ has measure zero.

## Formulas:

Consider that f and g are Riemann integrable on the interval $\left[c,d\right]$, then the following formulas hold:

• $\underset{c}{\overset{d}{\int }}\left(f+g\right)dx=\underset{c}{\overset{d}{\int }}fdx+\underset{c}{\overset{d}{\int }}gdx$
• $\underset{c}{\overset{d}{\int }}\left(f-g\right)dx=\underset{c}{\overset{d}{\int }}fdx-\underset{c}{\overset{d}{\int }}gdx$
• $\left|\underset{c}{\overset{d}{\int }}fdx\right|\le \underset{c}{\overset{d}{\int }}\left|f\right|dx$
• $\underset{c}{\overset{d}{\int }}\left(kf\right)dx\le k\underset{c}{\overset{d}{\int }}fdx$

Let g be a bounded function on $\left[c,d\right]$ and let $P=\left\{c={t}_{0}<{t}_{1}<{t}_{2}<\cdots {t}_{k-1}<{t}_{k}<\cdots {t}_{n}=d\right\}$ be a partition of $\left[c,d\right]$. A Riemann sum of g associated with the partition P is a sum of the form:

$\sum _{k=1}^{n}g\left({x}_{k}\right)\left({t}_{k}-{t}_{k-1}\right)$

Here ${x}_{k}\in \left[{t}_{k-1},{t}_{k}\right]$ for $k=1,2,\dots ,n$.

## Context and Applications

• It is used in everything from calculus to physics.
• It is used in partial differential equations and trigonometric series representation of functions.
• It is involved in the analysis of real-number sequences' limits and convergence.
• This is used in the study of real-valued functions' continuity and other properties.

## Practice Problem

Question: Using the properties of the Riemann integral, show that $\left|\underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)dx\right|\le \frac{16{\pi }^{3}}{3}$.

Solution: Recall the property of Riemann integral that deals with absolute value: $\left|\underset{c}{\overset{d}{\int }}fdx\right|\le \underset{c}{\overset{d}{\int }}\left|f\right|dx$

Now, consider the provided expression $\left|\underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)dx\right|$.

We have,

$\begin{array}{c}\left|\underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)dx\right|\le \underset{-2\pi }{\overset{2\pi }{\int }}\left|{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)\right|dx\\ \le \underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}\cdot 1dx\end{array}$

Because the absolute value of sine trigonometric function is 1, so $\left|\mathrm{sin}x\right|\le 1$.

Simplify it further as,

$\begin{array}{c}\left|\underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)dx\right|\le \underset{-2\pi }{\overset{2\pi }{\int }}\left|{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)\right|dx\\ \le \underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}\cdot 1dx\\ =2\underset{0}{\overset{2\pi }{\int }}{x}^{2}dx\end{array}$

Because ${x}^{2}$ is an even function.

Therefore, integrating the function,

$\begin{array}{c}\left|\underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)dx\right|=2{\left[\frac{{x}^{3}}{3}\right]}_{0}^{2\pi }\\ =\frac{2}{3}\left(8{\pi }^{3}\right)\\ =\frac{16{\pi }^{3}}{3}\end{array}$

Hence, we have proved that $\left|\underset{-2\pi }{\overset{2\pi }{\int }}{x}^{2}{\mathrm{sin}}^{8}\left({e}^{x}\right)dx\right|\le \frac{16{\pi }^{3}}{3}$ using the Riemann integrability concepts.

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