Lesson Explainer: Properties of Definite Integrals Mathematics • Higher Education

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 ๐‘“(๐‘ฅ) from ๐‘ฅ=๐‘Ž to ๐‘ฅ=๐‘ can be interpreted as the signed area under the curve of ๐‘“(๐‘ฅ) from ๐‘ฅ=๐‘Ž to ๐‘ฅ=๐‘; 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 ๐‘ฆ=๐‘“(๐‘ฅ), bounded by ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘, by first splitting up the interval [๐‘Ž,๐‘] into ๐‘› subintervals of equal width, [๐‘ฅ,๐‘ฅ]๏ƒ๏Šฑ๏Šง๏ƒ for ๐‘–=1,โ€ฆ,๐‘›, as shown in the diagram.

This gives ๐‘› rectangles of equal width, ฮ”๐‘ฅ, where the height of each rectangle is given by the value of the function at each point, ๐‘“(๐‘ฅ)๏ƒ, from the right endpoint of each subinterval. The area of each rectangle is the product of this height and width, ๐‘“(๐‘ฅ)ฮ”๐‘ฅ๏ƒ. We can estimate the area under the curve of ๐‘“(๐‘ฅ) by summing the areas of each rectangle as areaโ‰ˆ๐‘“(๐‘ฅ)ฮ”๐‘ฅ+โ‹ฏ+๐‘“(๐‘ฅ)ฮ”๐‘ฅ=๏„š๐‘“(๐‘ฅ)ฮ”๐‘ฅ.๏Šง๏Š๏Š๏ƒ๏Šฒ๏Šง๏ƒ

This is also known as the right Riemann sum. As the number of rectangles ๐‘› gets larger and the width ฮ”๐‘ฅ 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 ๐‘“ that is continuous and defined on the interval [๐‘Ž,๐‘], we can divide the interval into ๐‘› subintervals [๐‘ฅ,๐‘ฅ]๏ƒ๏Šฑ๏Šง๏ƒ of equal width, ฮ”๐‘ฅ, and choose sample points ๐‘ฅโˆˆ[๐‘ฅ,๐‘ฅ]โˆ—๏ƒ๏ƒ๏Šฑ๏Šง๏ƒ. The definite integral from ๐‘ฅ=๐‘Ž to ๐‘ฅ=๐‘ is defined in terms of the Riemann sum as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„š๐‘“๏€น๐‘ฅ๏…ฮ”๐‘ฅ,๏Œป๏Œบ๏Šโ†’โˆž๏Š๏ƒ๏Šฒ๏Šงโˆ—๏ƒdlim where ๐‘ฅ=๐‘Ž+๐‘–ฮ”๐‘ฅ,ฮ”๐‘ฅ=๐‘โˆ’๐‘Ž๐‘›=๐‘ฅโˆ’๐‘ฅ๏ƒ๏ƒ๏ƒ๏Šฑ๏Šง provided that the limit exists and gives the same value for all sample points ๐‘ฅโˆˆ[๐‘ฅ,๐‘ฅ]โˆ—๏ƒ๏ƒ๏Šฑ๏Šง๏ƒ.

It does not matter which sample point ๐‘ฅโˆ—๏ƒ in the subinterval [๐‘ฅ,๐‘ฅ]๏ƒ๏Šฑ๏Šง๏ƒ is taken to be. Since the difference or width of the summands ฮ”๐‘ฅโ†’0, so does the difference between any two points in the interval. This is because the choice of ๐‘ฅโˆ—๏ƒ 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 ๐‘ฅ=๐‘ฅโˆ—๏ƒ๏ƒ, that is, the function ๐‘“ 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 ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„š๐‘“(๐‘ฅ)ฮ”๐‘ฅ.๏Œป๏Œบ๏Šโ†’โˆž๏Š๏ƒ๏Šฒ๏Šง๏ƒdlim 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 ๐‘› equal-width rectangles and the diagram, with the limit as ๐‘›โ†’โˆž.
  • If ๐‘ฅ=๐‘ฅโˆ—๏ƒ๏ƒ๏Šฑ๏Šง, that is, the function ๐‘“ is evaluated at the left endpoint of each subinterval, then we have the left Riemann sum.
  • If ๐‘ฅ=(๐‘ฅ+๐‘ฅ)2โˆ—๏ƒ๏ƒ๏ƒ๏Šฑ๏Šง, that is, the function ๐‘“ 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 ๐‘ฅ-axis is always positive, while that below the ๐‘ฅ-axis is always negative, as shown in the diagram.

If there are parts of the curve that are below and above the ๐‘ฅ-axis in the interval [๐‘Ž,๐‘], then the definite integral will be the area above the ๐‘ฅ-axis minus the area below the ๐‘ฅ-axis, within the interval [๐‘Ž,๐‘].

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 (Newtonโ€“Leibniz Axiom)

Let ๐‘“ be a real-valued function on a closed interval [๐‘Ž,๐‘] and ๐น an antiderivative of ๐‘“ in [๐‘Ž,๐‘]: ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ).

If ๐‘“ is Riemann integrable on [๐‘Ž,๐‘], then ๏„ธ๐‘“(๐‘ก)๐‘ก=[๐น(๐‘ก)]=๐น(๐‘)โˆ’๐น(๐‘Ž).๏Œป๏Œบ๏Œป๏Œบd

By the first part of the theorem, antiderivatives of ๐‘“ always exist when ๐‘“ is continuous, but the second part of the fundamental theorem of calculus is somewhat stronger as it does not assume that ๐‘“ is continuous.

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

Properties of Definite Integrals

  1. The variable ๐‘ฅ that appears in definite integrals is called the dummy variable and we can replace this with another to get the same result: ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ก)๐‘ก.๏Œป๏Œบ๏Œป๏Œบdd
  2. The definite integral of a constant ๐‘โˆˆโ„ is proportional to the width of the interval: ๏„ธ๐‘๐‘ฅ=๐‘(๐‘โˆ’๐‘Ž).๏Œป๏Œบd
  3. The zero-width limit when ๐‘=๐‘Ž implies that ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=0.๏Œบ๏Œบd
  4. We can interchange the limits on any definite integral using ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œบ๏Œปdd
  5. We can split up definite integrals with a sum or difference: ๏„ธ(๐‘“(๐‘ฅ)ยฑ๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅยฑ๏„ธ๐‘”(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบddd
  6. We can factor out a constant ๐‘โˆˆโ„ from definite integrals: ๏„ธ๐‘๐‘“(๐‘ฅ)๐‘ฅ=๐‘๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบdd
  7. We can also split up the integral with the limits [๐‘Ž,๐‘] for some value ๐‘โˆˆโ„ over two adjacent intervals [๐‘Ž,๐‘] and [๐‘,๐‘]: ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผddd
  8. If ๐‘“(๐‘ฅ)โ‰ฅ0, then ๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ฅ0.๏Œป๏Œบd
  9. If ๐‘“(๐‘ฅ)โ‰ฅ๐‘”(๐‘ฅ), then ๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ฅ๏„ธ๐‘”(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบdd
  10. We also have the bounded property; if ๐‘šโ‰ค๐‘“(๐‘ฅ)โ‰ค๐‘€, then ๐‘š(๐‘โˆ’๐‘Ž)โ‰ค๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ค๐‘€(๐‘โˆ’๐‘Ž).๏Œป๏Œบd
  11. The modulus property is given by ||||๏„ธ๐‘“(๐‘ฅ)๐‘ฅ||||โ‰ค๏„ธ|๐‘“(๐‘ฅ)|๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบdd
  12. Finally, we have a property for even and odd functions when integrating over the interval [โˆ’๐‘Ž,๐‘Ž].
    For even functions ๐‘“(โˆ’๐‘ฅ)=๐‘“(๐‘ฅ), we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฑ๏Œบ๏Œบ๏Šฆdd For odd functions ๐‘“(โˆ’๐‘ฅ)=โˆ’๐‘“(๐‘ฅ), we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=0.๏Œบ๏Šฑ๏Œบd

The second property corresponds to the definite integral of the constant function ๐‘“(๐‘ฅ)=๐‘ between ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘, as shown in the diagram.

The definite integral gives the signed area under the curve equal to the area of the rectangle with lengths |๐‘| and |๐‘โˆ’๐‘Ž|, 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 ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=(๐น(๐‘)โˆ’๐น(๐‘Ž))+(๐น(๐‘)โˆ’๐น(๐‘))=๐น(๐‘)โˆ’๐น(๐‘Ž).๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผddd

This is an intuitive property, since the areas of each of the parts add up to the total area over [๐‘Ž,๐‘]. This can be visualized as follows.

When we are given a particular function to integrate over an interval [๐‘Ž,๐‘], 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, Letโ€™s evaluate the definite integral of ๐‘“(๐‘ฅ)=๐‘ฅ5+๐‘ฅ๏Šฉ๏Šจ from ๐‘ฅ=โˆ’7 to ๐‘ฅ=7, as shown in the plot.

The function ๐‘“ is odd since it satisfies ๐‘“(โˆ’๐‘ฅ)=โˆ’๐‘“(๐‘ฅ). Therefore, using the 11th property of definite integrals for odd functions, we have ๏„ธ๐‘ฅ5+๐‘ฅ๐‘ฅ=0.๏Šญ๏Šฑ๏Šญ๏Šฉ๏Šจd

This can also be realized from the plot of the graph, since the integral on the right side of the axis (๐‘ฅ>0) exactly cancels the integral on the left side (๐‘ฅ<0), since the latter below the ๐‘ฅ-axis will give a negative value.

Suppose we want to find the definite integral of ๐‘“(๐‘ฅ)=๐‘ฅ๐‘ฅ๏Šจcos from ๐‘ฅ=โˆ’๐œ‹ to ๐‘ฅ=๐œ‹, as shown in the plot.

If we are given the integral ๏„ธ๐‘ฅ๐‘ฅ๐‘ฅ=โˆ’2๐œ‹,๏Ž„๏Šฆ๏Šจcosd which we can find by using integration by parts twice, then we can evaluate the integral over the interval [โˆ’๐œ‹,๐œ‹] by using the fact that the function ๐‘“ is even since it satisfies ๐‘“(โˆ’๐‘ฅ)=๐‘“(๐‘ฅ). Therefore, using the 11th property of definite integrals for even functions, we have ๏„ธ๐‘ฅ๐‘ฅ๐‘ฅ=2๏„ธ๐‘ฅ๐‘ฅ๐‘ฅ=โˆ’4๐œ‹.๏Ž„๏Šฑ๏Ž„๏Šจ๏Ž„๏Šฆ๏Šจcosdcosd

This can also be realized from the plot of the graph, since the integral on the right side of the axis (๐‘ฅ>0) is exactly the same as the integral on the left side (๐‘ฅ<0).

Now, Letโ€™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 ๏„ธ๐‘”(๐‘ฅ)๐‘ฅ=10๏Šฎ๏Šฑ๏Šญd, determine the value of ๏„ธ7๐‘”(๐‘ฅ)๐‘ฅ๏Šฑ๏Šญ๏Šฎd.

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 ๏„ธ๐‘๐‘“(๐‘ฅ)๐‘ฅ=๐‘๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,๐‘โˆˆโ„,๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบ๏Œบ๏Œปdddd

On applying these properties, we can determine the value of the given integral as ๏„ธ7๐‘”(๐‘ฅ)๐‘ฅ=7๏„ธ๐‘”(๐‘ฅ)๐‘ฅ=โˆ’7๏„ธ๐‘”(๐‘ฅ)๐‘ฅ=โˆ’7ร—10=โˆ’70.๏Šฑ๏Šญ๏Šฎ๏Šฑ๏Šญ๏Šฎ๏Šฎ๏Šฑ๏Šญddd

Now, Letโ€™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 ๐‘“ is continuous on [โˆ’4,4] and satisfies ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=9๏Šช๏Šฆd. Determine ๏„ธ[๐‘“(๐‘ฅ)โˆ’6]๐‘ฅ๏Šช๏Šฆd.

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 ๏„ธ[๐‘“(๐‘ฅ)โˆ’๐‘”(๐‘ฅ)]๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ๐‘๐‘ฅ=๐‘(๐‘โˆ’๐‘Ž),๐‘โˆˆโ„.๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบdddd

On applying these properties, we can determine the value of the given integral as ๏„ธ[๐‘“(๐‘ฅ)โˆ’6]๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ6๐‘ฅ=9โˆ’6(4โˆ’0)=โˆ’15.๏Šช๏Šฆ๏Šช๏Šฆ๏Šช๏Šฆddd

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 ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=82๏Šซ๏Šฑ๏Šชd and ๏„ธ๐‘”(๐‘ฅ)๐‘ฅ=74๏Šซ๏Šฑ๏Šชd, find ๏„ธ[2๐‘“(๐‘ฅ)โˆ’4๐‘”(๐‘ฅ)]๐‘ฅ๏Šซ๏Šฑ๏Šชd.

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 ๏„ธ[๐‘“(๐‘ฅ)โˆ’๐‘”(๐‘ฅ)]๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ๐‘๐‘“(๐‘ฅ)๐‘ฅ=๐‘๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,๐‘โˆˆโ„.๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบddddd

On applying these properties, we can determine the value of the given integral as ๏„ธ[2๐‘“(๐‘ฅ)โˆ’4๐‘”(๐‘ฅ)]๐‘ฅ=๏„ธ2๐‘“(๐‘ฅ)๐‘ฅ+๏„ธโˆ’4๐‘”(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’4๏„ธ๐‘”(๐‘ฅ)๐‘ฅ=2ร—82โˆ’4ร—74=โˆ’132.๏Šซ๏Šฑ๏Šช๏Šซ๏Šฑ๏Šช๏Šซ๏Šฑ๏Šช๏Šซ๏Šฑ๏Šช๏Šซ๏Šฑ๏Šชddddd

Now, Letโ€™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 [โˆ’2,5], the values of ๐‘“ lie in the interval [๐‘š,๐‘€]. Between which bounds does ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šซ๏Šฑ๏Šจd lie?

Answer

In this example, we want to find the bounds of an integral using the property where the values of ๐‘“ lie in a particular interval.

In particular, the following property states that if ๐‘šโ‰ค๐‘“(๐‘ฅ)โ‰ค๐‘€, then ๐‘š(๐‘โˆ’๐‘Ž)โ‰ค๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ค๐‘€(๐‘โˆ’๐‘Ž).๏Œป๏Œบd

On applying this property with ๐‘Ž=โˆ’2 and ๐‘=5, we have ๐‘โˆ’๐‘Ž=7. Thus, 7๐‘šโ‰ค๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ค7๐‘€.๏Šซ๏Šฑ๏Šจd

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 ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’2.4๏Šจ๏Šฑ๏Šซd and ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’1.4๏Šฑ๏Šง๏Šฑ๏Šซd, find ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šจ๏Šฑ๏Šงd.

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 ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผddd

On applying this property and substituting the given values, we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,โˆ’2.4=โˆ’1.4+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,๏Šจ๏Šฑ๏Šซ๏Šฑ๏Šง๏Šฑ๏Šซ๏Šจ๏Šฑ๏Šง๏Šจ๏Šฑ๏Šงdddd which upon rearranging gives ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’2.4+1.4=โˆ’1.๏Šจ๏Šฑ๏Šงd

Now, Letโ€™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 ๐‘“ is continuous on โ„. Write ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šฉ๏Šฑ๏Šจ๏Šช๏Šฉ๏Šฆ๏Šฑ๏Šจddd in the form ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบd.

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 ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผ๏Œป๏Œบ๏Œบ๏Œปddddd

We can combine the first two terms using the first property as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šฉ๏Šฑ๏Šจ๏Šช๏Šฉ๏Šช๏Šฑ๏Šจddd and rewrite the third term as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Šฆ๏Šฑ๏Šจ๏Šฑ๏Šจ๏Šฆdd

Therefore, applying the first property again on the remaining two terms, we get ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Šฉ๏Šฑ๏Šจ๏Šช๏Šฉ๏Šฆ๏Šฑ๏Šจ๏Šฆ๏Šฑ๏Šจ๏Šฉ๏Šฑ๏Šจ๏Šช๏Šฉ๏Šฑ๏Šจ๏Šฆ๏Šช๏Šฑ๏Šจ๏Šช๏Šฆddddddddd

In the next example, we will use the property of definite integrals for even functions over an interval [โˆ’๐‘Ž,๐‘Ž] to evaluate the integral from a specified value.

Example 7: The Definite Integration of Even Functions

If the even function ๐‘“ is continuous over the interval [โˆ’4,4], where ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏Šช๏Šฆd, determine the value of ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šช๏Šฑ๏Šชd.

Answer

In this example, we want to evaluate a definite integral by using the property of the integral of even functions over an interval [โˆ’๐‘Ž,๐‘Ž].

The property of definite integrals we will make use of for even functions ๐‘“(โˆ’๐‘ฅ)=๐‘“(๐‘ฅ) is ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฑ๏Œบ๏Œบ๏Šฆdd

On applying this property, we can determine the value of the given integral as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=4.๏Šช๏Šฑ๏Šช๏Šช๏Šฆdd

Now, Letโ€™s look at an example where we will use the property of definite integrals for odd functions over an interval [โˆ’๐‘Ž,๐‘Ž] 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 ๐‘“ is odd, continuous on [โˆ’1,7], and satisfies ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’17๏Šญ๏Šงd. Determine ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šญ๏Šฑ๏Šงd.

Answer

In this example, we want to evaluate a definite integral by using the property of the integral of odd functions over an interval [โˆ’๐‘Ž,๐‘Ž] and splitting the integral between two adjacent intervals.

The properties of definite integrals we will make use of for odd functions ๐‘“(โˆ’๐‘ฅ)=โˆ’๐‘“(๐‘ฅ) are ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=0,๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฑ๏Œบ๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผdddd

On applying these properties, we can determine the value of the given integral as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=0โˆ’17=โˆ’17.๏Šญ๏Šฑ๏Šง๏Šง๏Šฑ๏Šง๏Šญ๏Šงddd

In the next example, we will use the property of definite integrals for even functions over an interval [โˆ’๐‘Ž,๐‘Ž] 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 ๐‘“ is even, continuous on [โˆ’8,8], and satisfies ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=19๏Šฎ๏Šฑ๏Šฎd and ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=13๏Šช๏Šฆd. Determine ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Šฑ๏Šช๏Šฑ๏Šฎd.

Answer

In this example, we want to evaluate a definite integral using the property of the integral of even functions over an interval [โˆ’๐‘Ž,๐‘Ž], 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 ๐‘“(โˆ’๐‘ฅ)=๐‘“(๐‘ฅ) are ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ,๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฑ๏Œบ๏Œบ๏Šฆ๏Œป๏Œบ๏Œบ๏Œป๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผddddddd

On applying the second and third property, we can write the definite integral as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Šฑ๏Šช๏Šฑ๏Šฎ๏Šฆ๏Šฑ๏Šฎ๏Šฑ๏Šช๏Šฆ๏Šฆ๏Šฑ๏Šฎ๏Šฆ๏Šฑ๏Šชddddd

We can determine each of these parts from the integrals given using the symmetry of even functions given as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฆ๏Šฆ๏Šฑ๏Œบdd

By applying this symmetry, we can determine the first part as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=13.๏Šฆ๏Šฑ๏Šช๏Šช๏Šฆdd

We can determine the second part in a similar way. By applying the first property, we note that ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=19,๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=192.๏Šฎ๏Šฑ๏Šฎ๏Šฎ๏Šฆ๏Šฎ๏Šฆddd

Thus, by symmetry, we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=192.๏Šฆ๏Šฑ๏Šฎd

Therefore, we can determine the given integral as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅโˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=192โˆ’13=โˆ’72.๏Šฑ๏Šช๏Šฑ๏Šฎ๏Šฆ๏Šฑ๏Šฎ๏Šฆ๏Šฑ๏Šชddd

Now, Letโ€™s look at an example where we will use the property of definite integrals for odd functions over an interval [โˆ’๐‘Ž,๐‘Ž] to evaluate the integral.

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

Determine ๏„ธ9๐‘ฅโˆ’6๐‘ฅ2๐‘ฅ+9๐‘ฅ๏Šง๏Šฑ๏Šง๏Šฉ๏Šจd.

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 [โˆ’1,1].

Letโ€™s first demonstrate that the integrand is indeed an odd function. Using ๐‘“(โˆ’๐‘ฅ)=โˆ’๐‘“(๐‘ฅ), we have ๐‘“(โˆ’๐‘ฅ)=9(โˆ’๐‘ฅ)โˆ’6(โˆ’๐‘ฅ)2(โˆ’๐‘ฅ)+9=โˆ’9๐‘ฅ+6๐‘ฅ2๐‘ฅ+9=โˆ’๏€พ9๐‘ฅโˆ’6๐‘ฅ2๐‘ฅ+9๏Š=โˆ’๐‘“(๐‘ฅ).๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ

The property of definite integrals we will make use of for odd functions is ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=0.๏Œบ๏Šฑ๏Œบd

On applying this property, we can evaluate the given integral as ๏„ธ9๐‘ฅโˆ’6๐‘ฅ2๐‘ฅ+9๐‘ฅ=0.๏Šง๏Šฑ๏Šง๏Šฉ๏Šจd

In the last example, we will use the property of definite integrals for even functions over an interval [โˆ’๐‘Ž,๐‘Ž] and the fundamental theorem of calculus to evaluate the definite integral.

Example 11: Evaluating the Definite Integration of a Power Function

Evaluate ๏„ธ๐‘ฅ๐‘ฅ๏Šง๏Šฑ๏Šง๏Šฎ๏Šฎd.

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 [โˆ’1,1].

Letโ€™s first demonstrate that the integrand is indeed an even function. Using the property of an even function, ๐‘“(โˆ’๐‘ฅ)=๐‘“(๐‘ฅ), we have ๐‘“(โˆ’๐‘ฅ)=(โˆ’๐‘ฅ)=๐‘ฅ=๐‘“(๐‘ฅ).๏Šฎ๏Šฎ๏Šฎ๏Šฎ

The property of definite integrals we will make use of for even functions is ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฑ๏Œบ๏Œบ๏Šฆdd

On applying this property on the given integral, we have ๏„ธ๐‘ฅ๐‘ฅ=2๏„ธ๐‘ฅ๐‘ฅ.๏Šง๏Šฑ๏Šง๏Šฎ๏Šฎ๏Šง๏Šฆ๏Šฎ๏Šฎdd

We begin by finding the antiderivative of ๐‘ฅ๏Šฎ๏Šฎ from its indefinite integral: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ89+๐ถ.๏Šฎ๏Šฎ๏Šฎ๏Šฏd

For the definite integral, we can apply the fundamental theorem of calculus that states that if ๐‘“ is continuous on [๐‘Ž,๐‘] and ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ), then ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=[๐น(๐‘ฅ)]=๐น(๐‘)โˆ’๐น(๐‘Ž).๏Œป๏Œบ๏Œป๏Œบd

We note that we can ignore the constant of integration for the antiderivative ๐น(๐‘ฅ), since this is canceled in the difference ๐น(๐‘)โˆ’๐น(๐‘Ž).

The integrand, for our integral, is clearly continuous and defined for all points ๐‘ฅโˆˆ[0,1]. Therefore, evaluating the antiderivative at the limit points and finding their difference, we have ๏„ธ๐‘ฅ๐‘ฅ=2๏„ธ๐‘ฅ๐‘ฅ=2๏–๐‘ฅ89๏ข=2๏€ฟ(1)89โˆ’(0)89๏‹=289.๏Šง๏Šฑ๏Šง๏Šฎ๏Šฎ๏Šง๏Šฆ๏Šฎ๏Šฎ๏Šฎ๏Šฏ๏Šง๏Šฆ๏Šฎ๏Šฏ๏Šฎ๏Šฏdd

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: ๏„ธ๐‘๐‘ฅ=๐‘(๐‘โˆ’๐‘Ž).๏Œป๏Œบd
  • Reversing the limits of integration, we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=โˆ’๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œบ๏Œปdd
  • Distributing the addition or difference over the integral, we have ๏„ธ(๐‘“(๐‘ฅ)ยฑ๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅยฑ๏„ธ๐‘”(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบ๏Œป๏Œบddd
  • Taking out a constant outside the integral, we have ๏„ธ๐‘๐‘“(๐‘ฅ)๐‘ฅ=๐‘๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบdd
  • Splitting the integral over two adjacent intervals, we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œผ๏Œบ๏Œป๏Œผddd
  • If ๐‘“(๐‘ฅ)โ‰ฅ0, then ๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ฅ0.๏Œป๏Œบd
  • If ๐‘“(๐‘ฅ)โ‰ฅ๐‘”(๐‘ฅ), then ๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ฅ๏„ธ๐‘”(๐‘ฅ)๐‘ฅ.๏Œป๏Œบ๏Œป๏Œบdd
  • If ๐‘šโ‰ค๐‘“(๐‘ฅ)โ‰ค๐‘€, then ๐‘š(๐‘โˆ’๐‘Ž)โ‰ค๏„ธ๐‘“(๐‘ฅ)๐‘ฅโ‰ค๐‘€(๐‘โˆ’๐‘Ž).๏Œป๏Œบd
  • For even functions ๐‘“(โˆ’๐‘ฅ)=๐‘“(๐‘ฅ), we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=2๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.๏Œบ๏Šฑ๏Œบ๏Œบ๏Šฆdd For odd functions ๐‘“(โˆ’๐‘ฅ)=โˆ’๐‘“(๐‘ฅ), we have ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=0.๏Œบ๏Šฑ๏Œบd

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