How to Calculate a Definite Integral?

ELet p(x) be the antiderivative of a continuous function f(x) defined on [a, b] then, the definite integral of f(x) over [a, b] is denoted by  and is equal to [p(b) – p(a)].

= P(b) – P(a)

The numbers a and b are called the limits of integration where a is called the lower limit and b is called the upper limit. The interval [a, b] is called the interval of the integration.

Note

• Constant integration is not included in the evaluation of the definite integral.
•  is read as “integral of f(x) from a to b”

Steps to find Definite Integrals

To find the definite integral of f(x) over interval [a, b] i.e.  we have following steps:

1. Find the indefinite integral ∫f(x)dx . 
2. Evaluate P(a) and P(b) where P(x) is antiderivative of f(x), P(a) is value of antiderivative at x=a and P(b) is value of antiderivative at x=b.
3. Calculate P(b) – P(a).
4. The resultant is the desired value of the definite integral. 

Definite Integrals by Substitution

For the integral . Let g(x) = t, then g'(x) dx = dt where for x = a , t = g(a) and for x = b, t = g(b).

If the variable is changed in the definite integral then substitution of a new variable affects the integrand, the differential (i.e. dx), and the limits.

The limits of the new variable t are the values of t corresponding to the values of the original variable x. It can be obtained by putting values of x in the substitution relation of x and t.  

Properties of Definite Integral

Property 1)

Proof:

Let p(x) be a antiderivative of f(x). Then,

{p(x)} = f(x) ⇒ \{p(z)} = f(z)

 = p(b) – p(a)                                               ——————- (i)

and  = p(b) – p(a)                                        ——————-(ii)

From (i) and (ii) 

Property 2)

If the limits of the definite integral are interchanged then, its value changes by a minus sign only. 

Proof:

Let p(x) be the antiderivative of f(x). Then,

 = p(b) – p(a)  

and  = -[p(a) – p(b)]  = p(b) – p(a)  

Property 3) where a < c < b

Proof:

Let p(x) be the antiderivative of f(x). Then,

 = p(b) – p(a)                                                                                               ——————(i)

 = [p(c) – p(a)] + [p(b) – p(c)] = p(b) – p(a)             ——————(ii)

From (i) and (ii)

Property 4)

Proof:

Let x = a – t . Then, dx = d(a – t) ⇒ dx = -dt

When x = 0 ⇒ t = a  and x = a ⇒ t = 0 

⇒                                                               [ By second property ]

⇒                                                              [ By first property ]

Property 5)

Proof:

Using third property

                                 ——————–(i)

Let x = – t , dx = -dt

Limits : x= -a  ⇒ t = a   and x = 0 ⇒ t = 0

      [By second property]

⇒                                                  [By first property] ———–(ii)

From (i) and (ii)

⇒

⇒ 

⇒ 

Property 6) If f(x) is a continuous function defined on [0, 2a],

Proof:

Using third property

                               —————–(i)

Consider  

Let x = 2a – t , dx = -d(2a – t) ⇒ dx = -dt

Limits : x= a  ⇒ t = a   and x = 2a  ⇒ t = 0

⇒                                          [ Using second property]

⇒                                         [ Using first property]

Substituting in (i)

Property 7)

Proof

Let t = a + b – x   ⇒ dt = -dx

Limits : x = a , y = b  and x = b , y = a

After putting value and limit of t in  

 ⇒

 ⇒                                         [Using second property]                 

 ⇒                                        [Using first property]

Solved Example on Definite Integrals

Problem 1: Evaluate: 

(i) 

(ii) 

(iii) 

Solution:

(i)  = 

                           = [2³ – 1³] 

                           = 8 – 1 

dx = 7

(ii)   = 

                            = (1/2)[log|-1| – log|-3| ] 

                            = (1/2)[ log 1 – log 3] 

                            = (1/2)[0 – log 3] 

  = (1/2)log 3

(iii) =  (sec² x – 1) dx 

                                    = 

                                    = [tan(π/4) – (π/4)] – [tan 0 – 0 ] 

= 1 – (π/4)

Problem 2: Evaluate:

Solution:

Let 5x² + 1 = t. Then, d(5x² + 1) = dt ⇒ 10 x dx = dt

 For limits : Lower limit ⇒ x = 0 then t = 5x² +1 = 1  and Upper limit ⇒ x = 1 then t = 5x² + 1 = 6

                   = 

                   = 

                   = (1/5) [log 6 – log 1]

  = (1/5) log 6

Problem 3: Evaluate :  

Solution:

        [Using definition of f(x)]

                                         =  [0 – ( -1 – 1)] +  [(1 + 1) – (0)]

Problem 4: Evaluate: 

Solution:

  ⇒ 

  ⇒ 

                        = 1 + 1

Problems 5: Evaluate: 

Solution:

I =          ———————(i)

I = 

Using 

I =           ——————-(ii)

Adding (i) and (ii) 

2I = 

2I = 

2I =  

2I =  

I = 0

Problem 6 : Evaluate : 

Solution:

I =                  —————–(i)

Using property 

I = 

I =         —————(ii)

Adding (i) and (ii)

2I = 

2I = 

2I  = 2 – 1

2I = 1

I = 1/2

FAQs on Definite Integrals

Question 1: What is meant by definite integrals?

Answer:

Definite integrals are integrals that are defined under proper limits i.e. their upper and lower limits are specified. It is represented as ∫_b^a f(x) dx where a is the upper limit and b is the lower limit of integration.

Question 2: How are definite integrals simplified?

Answer:

For simplifying definite integrals use the following steps:

• Simplify the integral normally.
• Substitute upper and lower limits to the answer of integration.
• Subtract both the answer obtained in step 2

Question 3: Write the formula for solving definite integrals.

Answer:

Suppose a definite integral of a function f(x) in the interval [a, b] is required, then,

∫_b^a f(x) dx = F(a) – F(b)  

where, ∫ f(x) dx = F(x) + C

Question 4: What does the value obtained from solving definite integral represent? Can it be negative?

Answer:

The value obtained from solving the definite integral represents the area. Yes, it can also be negative.