Properties of Definite Integrals Explained with Proofs and Examples

We will be learning some of the vital properties of definite integrals and the derivation of the proofs in this article to get an in-depth understanding of this concept. Integration is the estimation of an integral. It is just the opposite process of differentiation.  The integral maths concepts are used to find out the value of quantities like displacement, volume, area, and many more.  There are two types of Integrals namely, definite integral and indefinite integral. In this article, we will learn about definite integrals and their properties, which will help to solve integration problems based on them. 

Definite Integral Definition

An integral is known as a definite integral if and only if it has upper and lower limits. In Mathematics, there are many definite integral formulas and properties that are used frequently. To find the value of a definite integral, you have to find the difference between the values of the integral at the specified upper and lower limit of the independent variable and it is denoted as:

\[\int_{x}^{y}\]dx

Given below is a list of all the basic properties of the definite integral. This helps you while revising some properties of definite integrals with examples easily.

Here are the properties of definite integrals for even and odd functions. With these properties, you can solve the definite integral properties problems. 

Properties of Definite Integrals

Proofs of Definite Integrals Proofs

Property 1:   \[\int_{j}^{k}\]f(x)dx = \[\int_{j}^{k}\]f(t)dt

A simple property where you will have to only replace the alphabet x with t.

Property 2: \[\int_{j}^{k}\]f(x)g(x) = -\[\int_{j}^{k}\] f(x)g(x) , also \[\int_{k}^{j}\]f(x)g(x) = 0

Consider, m = \[\int_{j}^{k}\]f(x)g(x)

If the anti-derivative of f is f’, the second fundamental theorem of calculus is applied in order to get m = f’ ( k ) - f’ ( j ) = - f′( j ) - f′( k ) = \[\int_{j}^{k}\]xdx

Also, if j = k, then m = f’ ( k ) - f’ ( j ) = - f′( j ) - f′( j ) = 0. Therefore,  

\[\int_{k}^{j}\]f(x)g(x) = 0

Property 3: \[\int_{j}^{k}\]f(x)dx = \[\int_{j}^{l}\]f(x)dx + \[\int_{l}^{k}\]f(x)dx

If the anti-derivative of f is f’, the second fundamental theorem of calculus is applied in order to get 

\[\int_{j}^{k}\]f(x)dx = f’ ( k ) - f’ ( j ) . . . . . ( 1 )  

\[\int_{j}^{l}\]f(x)dx = f’ ( l ) - f’ ( j ) . . . . . ( 2 )  

\[\int_{l}^{k}\]f(x)dx = f’ ( k ) - f’ ( l ) . . . . . ( 3 )  

Adding equation ( 2) and ( 3 ), you get:

\[\int_{j}^{l}\]f(x)dx + \[\int_{l}^{k}\]f(x)dx = f’ ( l ) - f’ ( j ) + f’ ( k ) - f’ ( l ) = f’ ( k ) - f’ ( k ) = \[\int_{j}^{k}\]f(x)dx

Property 4: \[\int_{j}^{k}\]f(x)g(x) = \[\int_{j}^{k}\]f(j + k - x)g(x)

Let, m = ( j + k - x ), or x = ( j + k – m), so that dt = – dx … (4)

Also, note that when x = j, m = k and when x = k, m = j. So, \[\int_{j}^{k}\] wil be replaced by \[\int_{k}^{j}\]when we replace x by m. Therefore, \[\int_{j}^{k}\]f(x)dx = - \[\int_{j}^{k}\]f ( j + k - m ) dm … from equation (4)

From property 2, we know that \[\int_{j}^{k}\]f ( x ) dx = - \[\int_{j}^{k}\] f ( x ) dx. Use this property, to get 

\[\int_{j}^{k}\]f ( x ) dx = - \[\int_{j}^{k}\]f ( j + k - m ) dx

Now use property 1 to get \[\int_{j}^{k}\]f ( x ) dx = \[\int_{j}^{k}\]f ( j + k – x ) dx

Property 5:  \[\int_{0}^{k}\]f(x)g(x) = \[\int_{j}^{k}\]f(k - x)g(x)

Let, m = ( j - m ) or x = ( k – m ), so that dm = – dx…(5) Also, observe that when x = 0, m = j and when x = j, m = 0. So, \[\int_{0}^{j}\]will be replaced by \[\int_{0}^{j}\]when we replace x by m. Therefore,

\[\int_{0}^{j}\]f ( x ) dx = - \[\int_{0}^{j}\]f ( j - m ) dx from equation ( 5 )

From Property 2, we know that \[\int_{j}^{k}\]f ( x ) dx = - \[\int_{j}^{k}\]f ( x ) dx. Using this property , we get

\[\int_{0}^{j}\]f(x)dx = \[\int_{0}^{j}\]f ( j - m ) dm

Next, using Property 1, we get \[\int_{0}^{j}\]f ( x ) dx = \[\int_{0}^{j}\]f( j - x ) dx

Property 6: \[\int_{0}^{2k}\]f(x)dx = \[\int_{0}^{k}\]f(x)dx + \[\int_{0}^{k}\]f(2k - x)dx.....If f(2k - x) = f(x)

 From property 3, we know that

\[\int_{j}^{k}\]f(x)g(x) = - \[\int_{j}^{l}\]f(x)g(x), also , \[\int_{k}^{l}\]f(x)g(x) = 0 

Therefore, by applying this property to \[\int_{0}^{2k}\]f(x)dx , we got

\[\int_{0}^{2k}\]f(x)dx = \[\int_{0}^{k}\]f(x)dx + \[\int_{k}^{2k}\]f(x)dx , and after assuming \[\int_{0}^{k}\]f(x)dx = L1 and \[\int_{k}^{2k}\]f(x)dx = L2

\[\int_{0}^{2k}\]f(x)dx =  L1 +  L2  …(1)

Now, letting, y = (2k – x) or x = (2p – y), so that dy = -dx

Also, note that when x = p, then y = p, but when x = 2k, y = 0. Hence, L2  can be written as 

L2 =  \[\int_{k}^{2k}\]f(x)dx  =  \[\int_{k}^{0}\]f(2k - y)dy , and 

From the Property 2, we know that \[\int_{j}^{k}\]f(x)g(x) = -\[\int_{j}^{k}\] f(x)g(x)

Using this property to the equation of L2, we get

L2 = - \[\int_{0}^{k}\]f(2k - y)dy 

Now, by using Property 1, we get

L2 = \[\int_{0}^{k}\]f(2k - x)dx , using this value of L2 in the equation (1)

\[\int_{0}^{2k}\]f(x)dx =  L1 + L2 = \[\int_{0}^{k}\]f(x)dx + \[\int_{0}^{k}\]f(2k - x)dx

Hence, proving the property 6 of the definite Integrals