Essential Properties of Definite Integrals for Quick Problem Solving
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
FAQs on Properties of Definite Integrals Explained
1. What are the main properties of definite integrals as per the CBSE Class 12 syllabus for 2025-26?
The core properties of definite integrals that are essential for solving problems are:
- P₀: \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(t) \, dt\)
- P₁: \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\). Also, \(\int_{a}^{a} f(x) \, dx = 0\)
- P₂: \(\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx\)
- P₃: \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx\)
- P₄: \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx\)
- P₅: \(\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a-x) \, dx\)
- P₆: \(\int_{0}^{2a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx\), if \(f(2a-x) = f(x)\) and 0 if \(f(2a-x) = -f(x)\).
- P₇: \(\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx\), if f is an even function, and 0 if f is an odd function.
2. What is a definite integral and how does it differ from an indefinite integral?
A definite integral represents the accumulated value of a function over a specific interval, defined by upper and lower limits. Its result is a single numerical value. For example, \(\int_{a}^{b} f(x) \, dx\) calculates the net area under the curve of f(x) from x=a to x=b. In contrast, an indefinite integral, like \(\int f(x) \, dx\), represents a family of functions (antiderivatives) and includes a constant of integration, 'C'. The key difference is that a definite integral gives a value, while an indefinite integral gives a function.
3. What are the necessary conditions for a function to be integrable over a certain interval?
For a definite integral \(\int_{a}^{b} f(x) \, dx\) to be evaluated, the primary condition is that the function f(x) must be continuous on the closed interval [a, b]. If the function has a finite number of jump discontinuities within the interval but is otherwise bounded, it can still be integrated. However, functions with infinite discontinuities within the interval [a, b] are not integrable in the standard sense.
4. What are some key applications of definite integrals in mathematics and physics?
Definite integrals have numerous real-world applications beyond just solving equations. Some important examples include:
- Area Calculation: Finding the area under a curve, between two curves, or of a region bounded by curves.
- Volume Calculation: Determining the volume of a 3D solid by rotating a curve around an axis (solids of revolution).
- Physics: Calculating total work done by a variable force, finding the centre of mass of an object, or determining the total distance travelled from a velocity function.
- Statistics: Finding the probability of a continuous random variable falling within a certain range.
5. How does the property for even and odd functions, \(\int_{-a}^{a} f(x) \, dx\), simplify integration problems?
This property is a powerful shortcut. If you identify the function f(x) as even or odd over a symmetric interval [-a, a], you can simplify the calculation significantly.
- If f(x) is an odd function (where f(-x) = -f(x)), the integral \(\int_{-a}^{a} f(x) \, dx\) is always zero. This is because the signed areas on either side of the y-axis cancel each other out.
- If f(x) is an even function (where f(-x) = f(x)), the area is symmetric. The integral can be simplified to \(2\int_{0}^{a} f(x) \, dx\), which often makes the calculation easier.
6. Why is the property of splitting an integral, \(\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\), so important?
This property is crucial for dealing with complex functions or regions. Its importance lies in its ability to break down a difficult problem into simpler parts. For instance, it is essential for integrating piecewise functions, which have different definitions over different parts of the interval. It also helps in calculating the area of regions that may need to be split, such as when finding the area between two curves that intersect within the integration interval.
7. Can we interchange the limits of a definite integral, and what is the effect?
Yes, the limits of a definite integral can be interchanged. According to the properties of definite integrals, when you swap the upper and lower limits, the value of the integral is negated. This is expressed by the property: \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\). This is a direct consequence of the Fundamental Theorem of Calculus, where F(b) - F(a) = - (F(a) - F(b)).
