Solved Examples on How to Calculate Definite Integral?

Example 1: Evaluate: 

(i) [Tex]\bold{\int\limits_{1}^{2}x^2 dx}       [/Tex]     

(ii) [Tex]\bold{\int\limits_{0}^{1}\frac{1}{(2x – 3)} dx}          [/Tex]       

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

Solution:

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

                           = [2³ – 1³] 

                           = 8 – 1 

 [Tex]\int\limits_{1}^{2}x^2 [/Tex]dx = 7

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

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

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

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

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

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

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

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

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

Example 2: Evaluate: [Tex]\bold{\int\limits_{0}^{1}\frac{2x}{5x^2 +1}}[/Tex]

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

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

                   = [Tex]\frac{1}{5}\int\limits_{1}^{6}\frac{1}{t}dt[/Tex]

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

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

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

Example 3: Evaluate :  [Tex]\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}[/Tex]

Solution:

 [Tex]f(x) = \begin{cases} 1 – 2x \hspace{0.2cm},  x \le 0\\ 1 + 2x \hspace{0.2cm}, x\ge 0\end{cases}[/Tex]

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

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

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

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

 [Tex]\int\limits_{-1}^{1} f(x) dx = 4  [/Tex]

Example 4: Evaluate: [Tex]\int\limits_0^\pi|cos x| dx[/Tex]

Solution:

 [Tex]|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}[/Tex]

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

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

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

                        = 1 + 1

 [Tex]\int\limits_0^\pi|cos x| dx = 2[/Tex]

Example 5: Evaluate: [Tex]\int\limits_0^{\pi/2}log tan\hspace{0.1cm}x\hspace{0.1cm}dx[/Tex]

Solution:

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

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

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

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

Adding (i) and (ii) 

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

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

2I =  [Tex]\int\limits_0^{\pi/2}log1.dx[/Tex]

2I =  [Tex]\int\limits_0^{\pi/2}0.dx[/Tex]

I = 0

Example 6 : Evaluate : [Tex]\int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx[/Tex]

Solution:

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

Using property [Tex]\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b – x)dx[/Tex]

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

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

Adding (i) and (ii)

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

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

2I  = 2 – 1

2I = 1

I = 1/2

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 calculation of definite integration.