One of the basic problem of mathematics in its beginning was the problem of measurement of lengths, areas and volumes. We know how to determine the areas of the simple geometric shapes, for instance, of the triangle, square, rectangle…

The problem is how to determine the area of the shapes who have more complex boundaries, such as the part of the plane bounded by the graph of the function. For this purpose, we will approximate a part of the plane by using the simpler geometric shapes whose areas we can easy calculate, for instance, rectangles.

Let $f: [a,b] \to \mathbb{R}$ be the function which is continuous and non-negative on the interval $[a,b]$. We will divide the interval $[a,b]$ into $n$ subintervals, $n \in \mathbb{N}$, with mesh points $a = x_0 < x_1 < \ldots < x_{n-1} < x_n =b$.

**Note**. A finite subset of real numbers $x_0, x_1, \ldots, x_n$ with the property above is called a **partition** of the interval $[a, b]$ denoted by $P = \{x_0, x_1, \ldots, x_n \}$.

These subintervals do not need to be of equal length. Over the each subinterval $[x_{k-1}, x_k], k = 1, \ldots , n$, we will place two rectangles, one below and another over the graph of the function. Let $m_k$ and $M_k$ be their heights, respectively. By adding areas of these rectangles we will obtain the lower Riemann sum $L(f; P)$ and** **the upper Riemann sum $U(f; P)$. The **lower Riemann sum** of the function $f$ relative to $P$ is

$$L(f; P) = m_1(x_1 – x_0) + m_2(x_2 – x_1) + \ldots + m_n(x_n – x_{n-1}) = \sum_{k = 1}^n m_k (x_k – x_{k-1}),$$

where $m_k = \inf_{x \in [x_{k-1}, x_k]}$,

and the **upper Riemann sum** is

$$U(f; P) = M_1(x_1 – x_0) + M_2(x_2 – x_1) + \ldots + M_n(x_n – x_{n-1}) = \sum_{k = 1}^n M_k (x_k – x_{k-1}),$$

where $M_k = \sup_{x \in [x_{k-1}, x_k]}$.

The area $A$ of the region under the graph of the function $f$ is blended between the lower and upper sum:

$$\sum_{k = 1}^n m_k (x_k – x_{k-1}) < A < \sum_{k = 1}^n M_k (x_k – x_{k-1}).$$

If we put $\Delta x_k = x_k – x_{k-1}$, then we have

$$ \sum_{k = 1}^n m_k \Delta x_k < A < \sum_{k = 1}^n M_k \Delta x_k$$

If we imagine that $n \to \infty$ as $\Delta x \to 0$, then the lower and upper sum will have the same limit $I$. This limit needs to be equal to the area under the graph of the function $f$. The number $I$ does not depends on the way of calculating the lower and upper sum, it is determined only by the function $f$ and is called the integral of the function $f$.

**Definition of the integral as the limit of a sum**

We will divide the interval $[a, b]$ into $n$ subintervals where $\Delta x_k = x_k – x_{k-1}, k =1, \ldots, n$ is the width of the $k$th subinterval. The points $a=x_0, x_1, \ldots, x_n =b$ are their endpoints and we choose $x_k^{*} \in [x_{k-1}, x_k], k= 1, \ldots n$. From the Riemann sum, $f(x_k^{*})$ is the height of $k$th rectangle and $\Delta x_k$ is its width.

Let $f:[a,b] \to \mathbb{R}$ be positive continuous function on $[a, b]$. The common limit $I$ of the lower and upper Riemann sum is called a **definite integral** of the function $f$ from $a$ to $b$ denoted as

$$I = \int_{a}^{b} f(x) d(x) = \lim_{n \to \infty}\sum_{k=1}^{n} f(x_k^{*}) \Delta x_k.$$

If this limit exists, then we say that the function $f$ is **integrable** on $[a, b]$.

The integral is equal to the area of the region under the graph of the function $f$ over the interval $[a, b]$.

**Riemann’s integral and integrability **

Let $\varphi(a, b)$ be the set of all partitions of the interval $[a, b]$. The **upper Riemann integral** of the function $f$ on the interval $[a, b]$ is defined as

$$U(f) = \sup_{P \in \varphi} U(f; P),$$

and the **lower Riemann integral** is defined by

$$L(f) = \inf_{P \in \varphi} L(f; P).$$

We say that the function $f:[a, b] \to \mathbb{R}$ is **Riemann integrable** if it is bounded on $[a, b]$ and if the lower and upper Riemann integrals are equal, that is

$$ L(f) = U(f).$$

Then common value of $L(f)$ and $U(f)$ is called a **definite** (**Riemann**) **integral** of the function $f$ on the interval $[a, b]$ denoted by

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

The symbol $\int$ is called the **integral sign**, f(x) is called the **integrand** and $x$ the **variable of integration**. The numbers $a$ and $b$ are called the** lower** and **upper limit** of the integral.

**Example 1**. We will calculate the area of the region under the graph of the function $f: \mathbb{R} \to \mathbb{R}, f(x) = x + 1$ over the interval $[0, a]$.

We will divide the interval $[0, a]$ into $n$ equal subintervals. The length of each subinterval is $\frac{a}{n}$.

With $a_n$ we will denote the area of the region that specify the rectangles inscribed under the graph of the function $f$. The area of this region is the lower sum.

$$a_n = \frac{a}{n} \left [\left(\frac{a}{n} +1 \right) + \left( \frac{2a}{n} +1 \right) + \ldots \left( \frac{(n-1)a}{n} +1 \right) \right ] $$

$$= \frac{a}{n} \left [ \frac{a}{n}+ \frac{2a}{n} + \ldots + \frac{a(n-1)}{n} + (n -1) \right ]$$

$$= \frac{a^2}{n^2} \left[ 1 + 2 + \ldots + (n-1) + \frac{n(n-1)}{a}\right ] $$

Since the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$, then

$$1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2}.$$

Now we have:

$$a_n = \frac{a^2}{n^2} \left [\frac{n(n-1)}{2} + \frac{n(n-1)}{a} \right] $$

$$=a^2 \cdot \frac{(n-1)}{2n} + a \cdot \frac{n-1}{n}.$$

Similarly, with $A_n$ we will denote the area of the region that specify the rectangles described above the graph of the function $f$. We will obtain the expression for the upper sum.

$$A_n = \frac{a}{n} \left [\left( \frac{a}{n} + 1 \right) + \left( \frac{2a}{n} + 1 \right) + \ldots + \left ( \frac{na}{n} + 1 \right) \right]$$

$$= \frac{a}{n} \left[ \frac{a}{n} + \frac{2a}{n} + \ldots \frac{na}{n} + n \right] $$

$$= \frac{a^2}{n^2}\left [1 + 2 + \ldots + n + \frac{n^2}{a} \right]$$

$$ = \frac{a^2}{n^2} \cdot \left[ \frac{n(n+1)}{2} + \frac{n^2}{a} \right] $$

$$= a^2 \cdot \frac{n + 1}{2n} + a.$$

The area $A$ of the region under the graph of the given function $f$ is blended between the lower and upper sum:

$$ a^2 \frac{(n-1)}{2n} + a \frac{n-1}{n} < A < a^2 \frac{n + 1}{2n} + a,$$

that is

$$ \left ( \frac{1}{2} a^2 + a \right ) \left ( 1 – \frac{1}{n} \right) < A < \frac{1}{2}a^2 \left ( 1 + \frac{1}{n} \right) + a.$$

The lower and upper sum have the same limit when $n \to \infty$ which is equal to $\frac{1}{2} a^2 + a$. Therefore,

$$A = \frac{1}{2} a^2 + a.$$

**Criteria for Riemann integrability**

Now we are going to specify some conditions for the existences of the Riemann integral.

**Theorem **(**Necessary and sufficient condition for Riemann integrability**). A function $f: [a, b] \to \mathbb{R}$ is Riemann integrable on $[a, b]$ if and only if $\forall \epsilon > 0$ exists a partition $P$ such that

$$U (f; P) – L(f; P) < \epsilon.$$

**Theorem**. If the function $f: [a, b] \to \mathbb{R}$ is a continuous function on $[a, b]$ then the function $f$ is integrable.

**Theorem**. If the function $f: [a, b] \to \mathbb{R}$ is a monotone and bounded function then $f$ is integrable.

**Properties of the integral**

1. ) **Monotonicity of the integral.**

Let $f, g: [a, b] \to \mathbb{R}$ be integrable functions and $f \le g$. Then

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

2.) **Linearity.** From the definition of the integral as the limit of integral sums it follows two of its properties which are called linearity of the integral:

a) Let $f:[a, b]\to \mathbb{R}$ be integrable function and $a \in \mathbb{R}$. Then $a \cdot f$ is integrable and

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

b) Let $f, g: [a, b] \to \mathbb{R}$ be integrable functions. Then $f + g$ is integrable and

$$\int_{a}^{b} [f(x) + g(x)] dx = \int_{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx.$$

3.) **Additivity**. Suppose that $f: [a, b] \to \mathbb{R}$ and $a < b < c$. Then the function $f$ is integrable on $[a, b]$ if and only if is integrable on $[a, c]$ and $[b, c]$ and

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

**Definition**. Let $f: [a, b] \to \mathbb{R}, a<b,$ be integrable function and $a \le c \le b$. Then

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

$$\int_{c}^{c} f(x) dx =0.$$

With the definition above the additivity property holds $\forall a, b, c \in \mathbb{R}$.

We defined the integral for non-negative functions. If we take the function $f$ which is negative on $[a, b]$, then the function $-f$ is positive on $[a, b]$ and the area under the graph of the function $-f$ is equal to the area above the graph of the function $f$.

If we need to calculate the area under the graph of the function on the interval inside which $f$ changes the sign, then we must divide this interval into subintervals by the zeroes of the function $f$ and use the additivity property of the integral.

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

**The first fundamental theorem of calculus **

A function $F: \left \langle a, b \right \rangle \to \mathbb{R}$ is called an antiderivative function of the function $f: \left \langle a, b \right \rangle \to \mathbb{R}$ if

$$F'(x) = f(x), \quad \forall x \in \left \langle a, b \right \rangle.$$

For instance, $F = x^4$ is an antiderivative of the function $f = 4x^3$ because

$$F'(x) = \left(x^4 \right)’ = 4x^3 = f(x).$$

Moreover, $F(x) = x^4 + 1$ is also an antiderivative of the function $f(x) = 4x^3$ because $(x^4 + 1)’ = 4x^3$.

More generally, if $F$ and $G$ are the antiderivatives of the same function $f$ then

$$F(x) = G(x) + C,$$

whereby $C$ is a constant.

**Theorem **(**The first fundamental theorem of calculus**). If $F: [a, b] \to \mathbb{R}$ is an antiderivative of the function $f$ which is integrable on $[a, b]$, then

$$\int_{a}^{b} f(x) dx = F(b) – F(a).$$

**Example 2**. Compute

$$\int_{1}^{2} x^3 dx$$

by using the first fundamental theorem of calculus.

**Solution**.

Firstly, we need to find an antiderivative $F(x)$ for $f(x) = x^3$. We take $F(x) = \frac{1}{4} x^3$, because $\left( \frac{1}{4} x^4 \right)’ = x^3 = f(x)$.

Now, by using the first fundamental of calculus, we have

$$\int_{1}^{2} x^3 dx = \int_{1}^{2} f(x) dx = F(2) – F(1) = \frac{1}{4} \cdot 2^4 – \frac{1}{4} \cdot 1^4 = 4 -1 = 3.$$

The process of calculating the definite integral we write in the following way:

$$ \int_{a}^{b} f(x) dx = F(x) \bigg|_{a}^{b} = F(b) – F(a).$$

**The second fundamental theorem of calculus**

We will observe a continuous function $f$. With $F(x)$ we will denote the following definite integral:

$$F(x) = \int_{a}^{x} f(t) dt.$$

The variable of integration is denoted with $t$, because the variable $x$ is the upper limit of the integral.

$F(x)$ is the area under the graph of the function $f$ over the interval $[a, x]$.

**Theorem **(**The second fundamental theorem of calculus**). If $f:[a, b] \to \mathbb{R}$ is continuous function on $[a,b]$ and $F(x) = \int_{a}^{x} f(t) dt$ then

$$F'(x) = f(x).$$

This means that the derivative of the definite integral, whose the upper limit is $x$, is equal to the integrand.