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How to Calculate a Definite Integral?

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  • Last Updated : 30 Aug, 2022
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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 \int\limits_{a}^{b}f(x)dx  and is equal to [p(b) – p(a)].

 \bold{\int\limits_{a}^{b}f(x)dx}   = 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.
  • \int\limits_{a}^{b}f(x)dx   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. \int\limits_{a}^{b}f(x)dx   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 \int_{a}^{b}f(g(x))g'(x)dx . Let g(x) = t, then g'(x) dx = dt where for x = a , t = g(a) and for x = b, t = g(b).

\int\limits_{a}^{b}f(g(x))g'(x)dx  = \int\limits_{g(a)}^{g(b)}f(t)dt

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) \bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(z)dz}

Proof:

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

\frac{d}{dx}  {p(x)} = f(x) ⇒ \\frac{d}{dz}    {p(z)} = f(z)

\int\limits_{a}^b f(x)dx = \big[p(x)\big]_a^b   = p(b) – p(a)                                               ——————- (i)

and \int\limits_{a}^b f(z)dz = \big[p(z)\big]_a^b   = p(b) – p(a)                                        ——————-(ii)

From (i) and (ii) 

\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(z)dz}

Property 2) \bold{\int\limits_{a}^{b}f(x)dx =-\int\limits_{b}^{a}f(x)dx}

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,

\int\limits_{a}^b f(x)dx   = p(b) – p(a)  

and -\int\limits_{b}^a f(x)dx   = -[p(a) – p(b)]  = p(b) – p(a)  

\bold{\int\limits_{a}^{b}f(x)dx =-\int\limits_{b}^{a}f(x)dx}

Property 3) \bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}               where a < c < b

Proof:

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

\int\limits_{a}^b f(x)dx   = p(b) – p(a)                                                                                               ——————(i)

\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx      = [p(c) – p(a)] + [p(b) – p(c)] = p(b) – p(a)             ——————(ii)

From (i) and (ii)

\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}

Property 4) \bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}

Proof:

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

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

\int\limits_{0}^{a}f(x)dx =-\int\limits_{a}^{0}f(a - t)dt

⇒ \int\limits_{0}^{a}f(x)dx =\int\limits_{0}^{a}f(a - t)dt                                                                [ By second property ]

⇒ \int\limits_{0}^{a}f(x)dx =\int\limits_{0}^{a}f(a - x)dx                                                               [ By first property ]

\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}

Property 5) \bold{\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx  & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}

Proof:

Using third property

\int\limits_{-a}^{a}f(x)dx=\int\limits_{-a}^{0}f(x)dx+\int\limits_{0}^{a}f(x)dx                                   ——————–(i)

Let x = – t , dx = -dt

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

\int\limits_{-a}^{0}f(x)dx=\int\limits_{a}^{0}f(-t)(-dt) =-\int\limits_{a}^{0}f(-t)dt=\int\limits_{0}^{a}f(-t)dt        [By second property]

 \int\limits_{-a}^{0}f(x)dx=\int\limits_{0}^{a}f(-x)dx                                                      [By first property] ———–(ii)

From (i) and (ii)

\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}f(-x)dx+\int\limits_{0}^{a}f(x)dx

\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}[f(-x)+f(x)]dx

⇒ \int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx  & , if f(-x) = f(x) \\ 0 & , if f(-x) = -f(x) \end{cases}

⇒ \bold{\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx  & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}

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

\bold{\int\limits_{0}^{2a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx  & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}

Proof:

Using third property

\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{a}^{2a}f(x)dx                                       —————–(i)

Consider  \int\limits_{a}^{2a}f(x)dx         

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

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

 \int\limits_{a}^{2a}f(x)dx=-\int\limits_{a}^{0}f(2a - t)dt

⇒  \int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - t)dt                                                   [ Using second property]

⇒  \int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - x)dx                                                  [ Using first property]

Substituting  \int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - x)dx          in (i)

\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{0}^{a}f(2a - x)dx = \int\limits_{0}^{a}[f(x) + f(2a - x)]dx

\bold{\int\limits_{0}^{2a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx  & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}

Property 7) \bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx}

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  \int\limits_{a}^{b}f(a + b - x)dx

 ⇒\int\limits_{a}^{b}f(a + b - x)dx =-\int\limits_{b}^{a}f(t)dt                  

 ⇒ \int\limits_{a}^{b}f(a + b - x)dx =\int\limits_{a}^{b}f(t)dt                                               [Using second property]                 

 ⇒ \int\limits_{a}^{b}f(a + b - x)dx =\int\limits_{a}^{b}f(x)dx                                              [Using first property]

\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx}

Solved Example on Definite Integrals

Problem 1: Evaluate: 

(i) \bold{\int\limits_{1}^{2}x^2 dx}            

(ii) \bold{\int\limits_{0}^{1}\frac{1}{(2x - 3)} dx}                 

(iii) \bold{\int\limits_{0}^{\pi/4}tan^2x dx}

Solution:

(i) \int\limits_{1}^{2}x^2dx      = \big[x^3\big]_{1}^{2}            

                           = [23 – 13

                           = 8 – 1 

 \int\limits_{1}^{2}x^2 dx = 7

(ii)  \int\limits_{0}^{1}\frac{1}{(2x - 3)} dx      \frac{1}{2}\big[log(2x-3)\big]_{0}^{1}            

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

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

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

\int\limits_{0}^{1}\frac{1}{(2x-3)} dx       = (1/2)log 3

(iii) \int\limits_{0}^{\pi/4}tan^2x dx     \int\limits_{0}^{\pi/4}             (sec2 x – 1) dx 

                                    = \big[tanx - x\big]_{0}^{\pi/4}            

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

 \int\limits_{0}^{\pi/4}tan^2x dx     = 1 – (π/4)

Problem 2: Evaluate: \bold{\int\limits_{0}^{1}\frac{2x}{5x^2 +1}}

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

\int\limits_{0}^{1}\frac{2x}{5x^2 +1}=\int\limits_{1}^{6}\frac{2x}{t}.\frac{dt}{10x}

                   = \frac{1}{5}\int\limits_{1}^{6}\frac{1}{t}dt

                   = \frac{1}{5}\big[log\hspace{0.1cm}t\big]_1^6

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

\int\limits_{0}^{1}\frac{2x}{5x^2 +1}         = (1/5) log 6

Problem 3: Evaluate :  \int\limits_{-1}^{1} \text {f(x)dx , where f(x) = }\begin{cases} 1 - 2x ,  x \le 0\\ 1 + 2x , x\ge 0\end{cases}

Solution:

 f(x) = \begin{cases} 1 - 2x \hspace{0.2cm},  x \le 0\\ 1 + 2x \hspace{0.2cm}, x\ge 0\end{cases}

 \int\limits_{-1}^{1} f(x) dx = \int\limits_{-1}^0 f(x)dx \hspace{0.2cm}+ \int\limits_{0}^1 f(x)dx

 \int\limits_{-1}^{1} f(x) dx = \int\limits_{-1}^0 (1-2x)dx \hspace{0.2cm}+ \int\limits_{0}^1 (1 + 2x)dx                [Using definition of f(x)]

\int\limits_{-1}^{1} f(x) dx = \big[x-x^2\big]_{-1}^0 \hspace{0.2cm}+  \big[x + x^2\big]_{0}^1

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

 \int\limits_{-1}^{1} f(x) dx = 4

Problem 4: Evaluate: \int\limits_0^\pi|cos x| dx

Solution:

 |cos x| =\begin{cases} cosx\hspace{0.4cm},  0\le x\le \pi/2\\ -cosx \hspace{0.2cm}, \pi/2\le x\le \pi\end{cases}

 \int\limits_0^\pi|cos x| dx = \int\limits_0^{\pi/2}|cos x| dx + \int\limits_{\pi/2}^\pi|cos x| dx

  ⇒ \int\limits_0^\pi|cos x| dx = \int\limits_0^{\pi/2}|cos x| dx + \int\limits_{\pi/2}^\pi (-cos x) dx

  ⇒ \int\limits_0^\pi|cos x| dx = \big[cos x\big]_0^{\pi/2} - [sin x\big] _{\pi/2}^\pi

                        = 1 + 1

 \int\limits_0^\pi|cos x| dx = 2

Problems 5: Evaluate: \int\limits_0^{\pi/2}log tan\hspace{0.1cm}x\hspace{0.1cm}dx

Solution:

I = \int\limits_0^{\pi/2}log tan\hspace{0.1cm}x\hspace{0.1cm}dx                  ———————(i)

I = \int\limits_0^{\pi/2}log tan\hspace{0.1cm}(\frac{\pi}{2}-x)\hspace{0.1cm}dx

Using \int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx

I = \int\limits_0^{\pi/2}log \hspace{0.1cm}cotx\hspace{0.1cm}dx                   ——————-(ii)

Adding (i) and (ii) 

2I = \int\limits_0^{\pi/2}[log \hspace{0.1cm}tanx\hspace{0.1cm} +log \hspace{0.1cm}cotx\hspace{0.1cm}]dx

2I = \int\limits_0^{\pi/2}[log \hspace{0.1cm}(tanx\hspace{0.1cm} cotx)\hspace{0.1cm}]dx

2I =  \int\limits_0^{\pi/2}log1.dx

2I =  \int\limits_0^{\pi/2}0.dx

I = 0

Problem 6 : Evaluate : \int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx

Solution:

I = \int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx                                 —————–(i)

Using property \int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx

I = \int\limits_1^2\frac{\sqrt {3-x}}{\sqrt{3-(3-x)}\hspace{0.1cm}+\sqrt{3-x}}dx

I =  \int\limits_1^2\frac{\sqrt {3-x}}{\sqrt{x}\hspace{0.1cm}+\sqrt{3-x}}dx                          —————(ii)

Adding (i) and (ii)

2I =  \int\limits_1^2\frac{\sqrt{x}+\sqrt {3-x}}{\sqrt{x}\hspace{0.1cm}+\sqrt{3-x}}dx

2I = \int\limits_1^21.dx =\big[x\big]_1^2        

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 ∫ba 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,

ba 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.


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