[Maths Class Notes] on Definite Integral Formulas Pdf for Exam

Concept of Definite Integrals

The definite integral is closely linked to the antiderivative and indefinite integral of a given function. The introduction of the concept of a  definite integral of a given function initiates with a function f(x) which is continuous on a closed interval (a,b). The interval which is given is divided into “n” subinterval is that, although not mandatory can be considered of equal lengths(Δx). An arbitrary domain value xi is selected in each subinterval and its corresponding subinterval length is calculated and these ‘n’ products are added to calculate their sums. The sum is known as Riemann sum and may be either positive, negative or zero relies upon the behavior of the function on a closed interval.

For example, if (fx) is greater than 0 on [a,b] then the Riemann sum will be the positive real number and if (fx) is lesser than 0 on [a,b], then the Riemann sum will be the negative real number. The Riemann sum of the function f( x) on [ a, b] is represented as as

Sn = f(x1) Δx + f(x2)Δx+ f(x3) Δx+…. + f(xn) Δx

or Sn = [sum_{i=1}^{n}] (fxi ) Δx

What is Integration in Maths?

In Maths, integration is a process of summing up parts to determine the whole. It is a reverse process of differentiation where we can split the function into parts. Integration is used to determine the summation under a very large scale. Calculating a total of small numbers is an easy task and can be done even manually, but calculating a total of large numbers where the limit could reach infinity is a complex task. In such a case, an integration method is used. Integration and Differentiation both are important concepts of calculus. There are two different types of integration namely:

• Definite Integral

• Indefinite integral 

This article delivers information about the concepts of definite integrals, definite integrals equations, properties of definite integrals, definite integration by parts formula, reduction formulas in definite integration etc.

Definite Integral Equation

An integral including both upper and lower limits is considered as a definite integral. A Riemann integral is considered as a definite integral where x is confined to fall on the real line.

[int_{a}^{infty}] f(x) dx = [lim_{brightarrowinfty}] [int_{a}^{b} f(x) dx]

[int_{a}^{b}] f(x)  dx = F(b) – F(a)

In the above definite integral equation a,∞, and b are  determined as the lower and upper limits, F(a) is considered as the lower limit value of the integral and F(b) is considered as the upper limit value of the integral

Definite Integration by Parts Formula

Definite integration by parts formula is generally used to integrate the product of two functions. The definite integration by parts formula is given as :

[int] p q dx = p [int] q dx – [int] p’ ([int] q dx ) dx

In the above Definite integration by parts formula.

p represents the function p(x)

q represents the function q (x)

p’ is derivative of the function p(x).

Reduction Formula in Definite Integration

Some of the reduction formulas in definite integration are:

• Reduction formula for sin – Sinn x dx = -1/n cos x sinn-1 x + n-1/n [int] sinn-2 x dx

• Reduction formula for cos = Cosn x dx = -1/n sin x cosn-1 x + n-1/n [int] cosn-2 x dx

• Reduction formula for natural logarithm – [int] (In x)n dx = x(In x)n – n [int] (In x)n-1 dx

Properties of Definite Integrals

Some of the properties of definite integrals are given below:

1. [int_{a}^{b}] f (x) dx = – [int_{b}^{a}] f (x) dx 

2. [int_{a}^{a}] f (x) dx = 0

3. [int_{a}^{b}]k f (x) dx = k[int_{a}^{b}] f (x) dx 

4. [int_{a}^{b}] f (x) ± g(x) dx = [int_{a}^{b}] f (x) dx ± [int_{a}^{b}] g(x) dx

5. [int_{a}^{b}] f (x) dx = [int_{a}^{c}] f (x) dx + [int_{c}^{b}] f (x) dx

6. [int_{a}^{b}] f (x) dx = [int_{a}^{b}]f (t) dt

Definite Integral Solved Examples of Definite Integral Formulas

Some of the solved examples of definite integrals are given below:

1. Given that, [int_{0}^{3}] x² dx = 8 , solve [int_{0}^{3}] 4x² dx

Solution: 

[int_{0}^{3}] 4x² dx =  4 [int_{0}^{3}]x² dx

= ( 4) × 8

= 32

2. Solve [int_{2}^{6}] 3 dx

Solution:

[int_{2}^{6}]  3 dx = 3(6-2)

= 12

3. Evaluate [int_{1}^{2}] xdx/(x² + 2)³

Solution:

 By using the substitution method, we get 

u = x² + 2

du = 2x dx

½ du = x dx

The limit of integration can be transformed from x values to their corresponding  values. When x= 1,u = 3 and when x =2 , u = 6, find 

[int_{1}^{2}] xdx/(x² + 2)³ = ½ [int_{3}^{6}] du/u³ 

= ½ [int_{3}^{6}] u-3 du

= ½ [ -½ u-2 ] [int_{3}^{6}]

= -¼ [(6)-2 – (3)-2]

= -¼ (1/36 – 1/9)

= 1/48

It is important to note that the substitution method is used to calculate definite integrals and it is not necessary to return back to the original variable if the limit of integration is transformed to the new variable values.’

Quiz Time

1. [int_{0}^{10y}] |sin y| dy is

1. 20

2. 8

3. 10

4. 18

2. [int_{-1.2}^{1/2}] cos y In ( 1+ y/1-y)dy is Equal to

1. 1

2. 2

3. 3

4. None of these