How to Calculate a Definite Integral?

A definite integral is a mathematical concept used in calculus to calculate the total effect of a function over a given time frame. It represents the net area between the function's graph and the x-axis during a certain time frame. A definite integral is calculated by determining the area under a curve between two specified locations on the x-axis. The definite integral is the net accumulation of the function's values over the specified interval. This article talks about how to calculate definite integral and also provide some solved examples based on the calculation of definite integration.

What is Definite Integral?

Let 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 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"

How 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:

Step 1: Find the indefinite integral ∫f(x)dx . 

Step 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.

Step 3: Calculate P(b) - P(a).

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  &amp; \text{, if $n$ is even} \\ 0 &amp; \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  &amp; , if f(-x) = f(x) \\ 0 &amp; , 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  &amp; \text{, if $n$ is even} \\ 0 &amp; \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  &amp; , if f(2a - x) = f(x) \\ 0 &amp; , 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  &amp; , if f(2a - x) = f(x) \\ 0 &amp; , 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}

Also, Check

• Calculus in Maths
• Integral Calculus
• Definite Integrals of Piecewise Functions
• Applications of Definite Integrals

Solved Examples on How to Calculate Definite Integral?

Example 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)

Example 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

Example 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  

Example 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

Example 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

Example 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