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 be the function which is continuous and non-negative on the interval . We will divide the interval into subintervals, , with mesh points .

**Note**. A finite subset of real numbers with the property above is called a **partition** of the interval denoted by .

These subintervals do not need to be of equal length. Over the each subinterval , we will place two rectangles, one below and another over the graph of the function. Let and be their heights, respectively. By adding areas of these rectangles we will obtain the lower Riemann sum and** **the upper Riemann sum . The **lower Riemann sum** of the function relative to is

where ,

and the **upper Riemann sum** is

where .

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

If we put , then we have

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

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

We will divide the interval into subintervals where is the width of the th subinterval. The points are their endpoints and we choose . From the Riemann sum, is the height of th rectangle and is its width.

Let be positive continuous function on . The common limit of the lower and upper Riemann sum is called a **definite integral** of the function from to denoted as

If this limit exists, then we say that the function is **integrable** on .

The integral is equal to the area of the region under the graph of the function over the interval .

**Riemann’s integral and integrability **

Let be the set of all partitions of the interval . The **upper Riemann integral** of the function on the interval is defined as

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

We say that the function is **Riemann integrable** if it is bounded on and if the lower and upper Riemann integrals are equal, that is

Then common value of and is called a **definite** (**Riemann**) **integral** of the function on the interval denoted by

The symbol is called the **integral sign**, f(x) is called the **integrand** and the **variable of integration**. The numbers and 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 over the interval .

We will divide the interval into equal subintervals. The length of each subinterval is .

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

Since the sum of the first natural numbers is , then

Now we have:

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

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

that is

The lower and upper sum have the same limit when which is equal to . Therefore,

**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 is Riemann integrable on if and only if exists a partition such that

**Theorem**. If the function is a continuous function on then the function is integrable.

**Theorem**. If the function is a monotone and bounded function then is integrable.

**Properties of the integral**

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

Let be integrable functions and . Then

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 be integrable function and . Then is integrable and

b) Let be integrable functions. Then is integrable and

3.) **Additivity**. Suppose that and . Then the function is integrable on if and only if is integrable on and and

**Definition**. Let be integrable function and . Then

With the definition above the additivity property holds .

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

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

**The first fundamental theorem of calculus **

A function is called an antiderivative function of the function if

For instance, is an antiderivative of the function because

Moreover, is also an antiderivative of the function because .

More generally, if and are the antiderivatives of the same function then

whereby is a constant.

**Theorem **(**The first fundamental theorem of calculus**). If is an antiderivative of the function which is integrable on , then

**Example 2**. Compute

by using the first fundamental theorem of calculus.

**Solution**.

Firstly, we need to find an antiderivative for . We take , because .

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

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

**The second fundamental theorem of calculus**

We will observe a continuous function . With we will denote the following definite integral:

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

is the area under the graph of the function over the interval .

**Theorem **(**The second fundamental theorem of calculus**). If is continuous function on and then

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