In this explainer, we will learn how to use properties of definite integration, such as the order of integration limits, zero-width limits, sums, and differences.

Definite integrals are all about the accumulation or sum of a particular quantity and are closely related to antiderivatives. They provide us with a powerful tool to help us understand and model real-world phenomena, as they appear in many disciplines from pure mathematics, with geometric applications such as surface area and volume, to physics when determining the mass of an object, the work done, or the pressure exerted on an object, to name just a few.

The definite integral of the function from to can be interpreted as the *signed* area under the curve of from to ; a visual representation of this integral is given in the diagram.

So, how are definite integrals defined? Before we give the precise definition, we note that we can estimate the area under the curve for some function , bounded by and , by first splitting up the interval into subintervals of equal width, for , as shown in the diagram.

This gives rectangles of equal width, , where the height of each rectangle is given by the value of the function at each point, , from the right endpoint of each subinterval. The area of each rectangle is the product of this height and width, . We can estimate the area under the curve of by summing the areas of each rectangle as

This is also known as the right Riemann sum. As the number of rectangles gets larger and the width gets smaller, this estimate will get closer to the true area under the curve. In fact, the definite integral, which gives the exact area under the curve, is defined by taking the limit of this sum as the number of rectangles approaches infinity.

### Definition: The Definite Integral

Given a function that is continuous and defined on the interval , we can divide the interval into subintervals of equal width, , and choose sample points . The definite integral from to is defined in terms of the Riemann sum as where provided that the limit exists and gives the same value for all sample points .

It does not matter which sample point in the subinterval is taken to be. Since the difference or width of the summands , so does the difference between any two points in the interval. This is because the choice of is arbitrary, which may produce different Riemann sums that converge to the same value. In particular, the common choices are given by the following:

- If , that is, the function is evaluated at the the right endpoint of each subinterval, then we have the right Riemann sum. The definite integral in terms of this sum is This is the choice that most people use when finding a specific Riemann sum or definite integral, for simplicity, and it corresponds to the example above with estimate of the area under the curve using equal-width rectangles and the diagram, with the limit as .
- If , that is, the function is evaluated at the left endpoint of each subinterval, then we have the left Riemann sum.
- If , that is, the function is evaluated at the midpoint of each subinterval, then we have the midpoint Riemann sum.

The definite integral always gives the signed area under the curve; the area given by the definite integral above the -axis is always positive, while that below the -axis is always negative, as shown in the diagram.

If there are parts of the curve that are below and above the -axis in the interval , then the definite integral will be the area above the -axis minus the area below the -axis, within the interval .

In practice, definite integrals are usually evaluated using the fundamental theorem of calculus by computing the antiderivative of the integrand and finding the difference evaluated at the limit points.

### The Second Part of the Fundamental Theorem of Calculus (Newton–Leibniz Axiom)

Let be a real-valued function on a closed interval and an antiderivative of in :

If is Riemann integrable on , then

By the first part of the theorem, antiderivatives of always exist when is continuous, but the second part of the fundamental theorem of calculus is somewhat stronger as it does not assume that is continuous.

Definite integrals also satisfy certain properties, similar to indefinite integrals, derivatives, and limits. Let’s recall these properties, which will be the focus of this explainer.

### Properties of Definite Integrals

- The variable that appears in definite integrals is called the dummy variable and we can replace this with another to get the same result:
- The definite integral of a constant is proportional to the width of the interval:
- The zero-width limit when implies that
- We can interchange the limits on any definite integral using
- We can split up definite integrals with a sum or difference:
- We can factor out a constant from definite integrals:
- We can also split up the integral with the limits for some value over two adjacent intervals and :
- If , then
- If , then
- We also have the bounded property; if , then
- The modulus property is given by
- Finally, we have a property for even and odd functions when integrating over the interval .

For even functions , we have For odd functions , we have

The second property corresponds to the definite integral of the constant function between and , as shown in the diagram.

The definite integral gives the signed area under the curve equal to the area of the rectangle with lengths and , up to a sign.

The other properties of definite integrals can be shown directly from the fundamental theorem of calculus; for instance, using the seventh property, we can split up the integral and apply the fundamental theorem of calculus to obtain

This is an intuitive property, since the areas of each of the parts add up to the total area over . This can be visualized as follows.

When we are given a particular function to integrate over an interval , we can use these properties to help evaluate the integral by simplifying it, often followed by an application of the fundamental theorem of calculus.

Now, Let’s evaluate the definite integral of from to , as shown in the plot.

The function is odd since it satisfies . Therefore, using the 11th property of definite integrals for odd functions, we have

This can also be realized from the plot of the graph, since the integral on the right side of the axis exactly cancels the integral on the left side , since the latter below the -axis will give a negative value.

Suppose we want to find the definite integral of from to , as shown in the plot.

If we are given the integral which we can find by using integration by parts twice, then we can evaluate the integral over the interval by using the fact that the function is even since it satisfies . Therefore, using the 11th property of definite integrals for even functions, we have

This can also be realized from the plot of the graph, since the integral on the right side of the axis is exactly the same as the integral on the left side .

Now, Let’s look at at a few examples to practice and deepen our understanding of the properties of definite integrals.

In the first example, we will evaluate a definite integral using the property of reversing limits of integration and factoring out a constant from an integral.

### Example 1: Evaluating a Definite Integral Using the Property of Reversing the Limits of Integration

If , determine the value of .

### Answer

In this example, we want to evaluate a definite integral using the property of reversing the limits of integration and factoring out a constant from the integral.

The properties of definite integrals we will make use of are

On applying these properties, we can determine the value of the given integral as

Now, Let’s look at example where we will evaluate a definite integral using the property of addition of two functions and the integral of a constant over the same interval.

### Example 2: Evaluating a Definite Integral Using the Property of Addition of the Integral of Two Functions over the Same Interval

The function is continuous on and satisfies . Determine .

### Answer

In this example, we want to evaluate a definite integral by using the property of addition of the integral of two functions and the integral of a constant over the same interval.

The properties of definite integrals we will make use of are

On applying these properties, we can determine the value of the given integral as

In the next example, we will evaluate a definite integral using the property of addition of two definite integrals and factoring out a constant from the integral.

### Example 3: Evaluating a Definite Integral Using the Property of Addition of Two Definite Integrals over the Same Interval

If and , find .

### Answer

In this example, we want to evaluate a definite integral by using the property of addition of the integral of two functions and factoring out a constant from the integral.

The properties of definite integrals we will make use of are

On applying these properties, we can determine the value of the given integral as

Now, Let’s look at an example where we will find the bounds of an integral where the values of the integrand lie in a particular interval.

### Example 4: Evaluating a Definite Integral Using the Bounded Property of Integrals

Suppose that on , the values of lie in the interval . Between which bounds does lie?

### Answer

In this example, we want to find the bounds of an integral using the property where the values of lie in a particular interval.

In particular, the following property states that if , then

On applying this property with and , we have . Thus,

In the next example, we will evaluate a definite integral using the property of splitting an integral over two adjacent intervals.

### Example 5: Evaluating a Definite Integral Using the Property of Addition of Two Definite Integrals over Two Adjacent Intervals

If and , find .

### Answer

In this example, we want to evaluate a definite integral using the property of splitting the integral between two adjacent intervals.

The property of definite integrals we will make use of is

On applying this property and substituting the given values, we have which upon rearranging gives

Now, Let’s look at an example where we will use the property of splitting an integral over adjecent intervals to express the sum of three definite integrals as one integral. We will also use a property to reverse the limits of integration.

### Example 6: Expressing the Sum of Three Definite Integrals over Adjacent Intervals as One Integral

The function is continuous on . Write in the form .

### Answer

In this example, we want to evaluate a definite integral using the property of reversing the limits of integration and splitting the integral between two adjacent intervals.

The properties of definite integrals we will make use of are

We can combine the first two terms using the first property as and rewrite the third term as

Therefore, applying the first property again on the remaining two terms, we get

In the next example, we will use the property of definite integrals for even functions over an interval to evaluate the integral from a specified value.

### Example 7: The Definite Integration of Even Functions

If the even function is continuous over the interval , where , determine the value of .

### Answer

In this example, we want to evaluate a definite integral by using the property of the integral of even functions over an interval .

The property of definite integrals we will make use of for even functions is

On applying this property, we can determine the value of the given integral as

Now, Let’s look at an example where we will use the property of definite integrals for odd functions over an interval to evaluate the integral from a specified value. We will also use the property to split the integral over two adjacent intervals.

### Example 8: Finding the Definite Integration of an Odd Function Using the Properties of Definite Integration

The function is odd, continuous on , and satisfies . Determine .

### Answer

In this example, we want to evaluate a definite integral by using the property of the integral of odd functions over an interval and splitting the integral between two adjacent intervals.

The properties of definite integrals we will make use of for odd functions are

On applying these properties, we can determine the value of the given integral as

In the next example, we will use the property of definite integrals for even functions over an interval to evaluate the integral from specified values. We will also use the property for reversing limits of integrals and splitting the integral over two adjacent intervals.

### Example 9: Finding the Definite Integration of an Even Function Using the Properties of Definite Integration

The function is even, continuous on , and satisfies and . Determine .

### Answer

In this example, we want to evaluate a definite integral using the property of the integral of even functions over an interval , reversing the limits of an integral, and splitting the integral between two adjacent intervals.

The properties of definite integrals we will make use of for even functions are

On applying the second and third property, we can write the definite integral as

We can determine each of these parts from the integrals given using the symmetry of even functions given as

By applying this symmetry, we can determine the first part as

We can determine the second part in a similar way. By applying the first property, we note that

Thus, by symmetry, we have

Therefore, we can determine the given integral as

Now, Let’s look at an example where we will use the property of definite integrals for odd functions over an interval to evaluate the integral.

### Example 10: Using the Properties of Definite Integration of an Odd Function to Evaluate an Integral

Determine .

### Answer

In this example, we want to evaluate a definite integral of an odd function, as shown in the graph, using the property of the integral of odd functions over an interval .

Let’s first demonstrate that the integrand is indeed an odd function. Using , we have

The property of definite integrals we will make use of for odd functions is

On applying this property, we can evaluate the given integral as

In the last example, we will use the property of definite integrals for even functions over an interval and the fundamental theorem of calculus to evaluate the definite integral.

### Example 11: Evaluating the Definite Integration of a Power Function

Evaluate .

### Answer

In this example, we want to evaluate a definite integral of an even function, as shown in the graph, using the property of the integral of even functions over an interval .

Let’s first demonstrate that the integrand is indeed an even function. Using the property of an even function, , we have

The property of definite integrals we will make use of for even functions is

On applying this property on the given integral, we have

We begin by finding the antiderivative of from its indefinite integral:

For the definite integral, we can apply the fundamental theorem of calculus that states that if is continuous on and , then

We note that we can ignore the constant of integration for the antiderivative , since this is canceled in the difference .

The integrand, for our integral, is clearly continuous and defined for all points . Therefore, evaluating the antiderivative at the limit points and finding their difference, we have

### Key Points

The key properties of definite integrals we should remember are as follows:

- The integral of a constant is proportional to the width of the interval:
- Reversing the limits of integration, we have
- Distributing the addition or difference over the integral, we have
- Taking out a constant outside the integral, we have
- Splitting the integral over two adjacent intervals, we have
- If , then
- If , then
- If , then
- For even functions , we have For odd functions , we have