PDF Chapter 07 Integration - Daffodil International University

.

Integration:

The process of finding an anti-derivative or integral of a function is called integration. It is the inverse process of differentiation. If ^{f x}

 

be a function of x related with another function ^{F x}

 

ⁱⁿ

such a way that

   

d F x f x

dx  then



^{f x dx}

 

^{F x}

 

^c

which is called an indefinite integral of ^{f x}

 

with respect to x, where ^{f x}

 

^{, F x}

 

^{and c} are called integrand, integral and constant of integration respectively.

And, ^b

     

a

f x dxF b F a



, which is called the definite integral of ^{f x}

 

^{from a to b, and ‘a’ is}

called the lower limit and ‘b’ the upper limit of the definite integral.

Fundamental Properties:

1.



f1

 

x  f2

 

x ... ... ...to nterms dx 



f1

 

x dx



f2

 

x dx... ...to nterms^. 2.



^{cf x dx}

 

^{c f x dx}

  

^{, where c} is a constant.

Integral Formula:

Problem: Solve the following integrations,

Chapter 07 Integration

1. ¹



¹



1

n

n x

x dx c where n

n

    





2.



dx x c

3. dx 2

x c

x  



4. dx ln x c

x  



5. sin



xdx  cosxc 6.



cosxdx sinxc 7. sec tan



x xdxsecxc

8. du.

uvdx u vdx vdx dx dx

 

   

 

   

.

(i) dx x x



^{(5 3 1)} (ii) e dx x



⁽₂^{1  9}x⁾ (iii) dx x

x



x^{2 2 1 (iv)}



^{(x2)(x3)2dx}

(v)



sin² xdx (vi)



tan²xdx (vii)



1 sin xdx Solution:

(i) Given that, ³ 1 (5x )dx

 x



Let, ³ 1

(5 )

I x dx





 x

3 1

5x dx dx









x

4

5 4

5 ln ln

4 4

x x c x x c

      (Ans.)

(ii) Given that, ( ^{1 9} ) 2

e x dx x 



Let, ( ^{1 9} )

2

I e x dx





x  1 9

2

dx e dxx





x 



9 9

(9 ) 9

x x

e e

x c x c

d x dx

     

(Ans.) (iii) Given that, dx

x x



x^{2 2 1} Let,

2 2 1

x x

I dx

x

 





2 1

( x 2 x )

x x x dx





 

3 1 1

2 2 2 2

x dx x dx x^ dx













5 3 1

2 2 2

5 3 1

2 2 2

x 2x x

   c

5 3 1

2 2 2

2 4

5x 3x 2x c

    (Ans.)

where c is an integrating constant.

(iv) Given that,



^{(x2)(x3)2dx}

Let, I 



(x2)(x3)²dx (x 2)(x2 6x 9)dx





  

3 2

(x 8x 21x 18)dx





  

4

3 2

8 21

4 3 2 18

x x x x c

     (Ans.)

where c is an integrating constant.

(v) Given that,



sin²xdx Let, I 



sin²xdx

¹ 2sin²

2 xdx





¹



^{1 cos 2}



2 x dx





 (Ans.)

1 sin 2

2 2

x x c

 

    (Ans.)

where c is an integrating constant.

(vi) Given that,



tan²xdx Let, I 



tan²xdx ^

 

^{sec2 x1}



^dx

^



^{tanx x}



^c. (Ans.)

where c is an integrating constant.

.

(vii) Given that,



1 sin xdx Let, I 



1 sin xdx

1 sin 2( ) 1 2sin( ) cos( )

2 2 2

x x x

dx dx





 





2 2

sin ( ) cos ( ) 2sin( ) cos( )

2 2 2 2

x x x x





  dx

2

sin cos

2 2

x x

     dx





       {sin( ) cos( )}

2 2

x x





 dx

2 cos( ) 2sin( )

2 2

x x

   c

2{sin( ) cos( )}

2 2

x x

  c

^(Ans.)

where c is an integrating constant.

Practice: Solve the following integrations, (i)



1 sin 2xdx (ii)

1 sin dx

 x



Method of Substitution:

Sometimes, the integration of given integral



^{f x dx}

 

is relatively difficult. In this case, we can replace x by 

 

^z and dx by ^'

 

^{z dz}for integrating easily. This process is known as method of substitution.

Problem: Solve the following integrations, (i)

1 2

sin 1

xdx x





 (ii)

1 ^x dx

e



Solution:

(i) Given that,

1 2

sin 1

xdx x





 Let,

1 2

sin 1

I x dx

x

 





(ii) Given that,

1 ^x dx

e



Let, 1 ^x

I dx

 e





.

Now, put ¹

2

sin 1

1 z x dz

dx x

   



1 2

dz dx

x

 

 Now I 



zdz

2

2 z c

 



^{sin 1}



²

2 x

c



 

where c is an integrating constant.

1 1

x x

dx e e

   

 





¹



x x

e dx

e



 







¹



x x

e dx

e





  





^{ I ln}



^{ex 1}



^c

where c is an integrating constant.

Exercise: Solve the following integrations, (i)

tan1

1 2

m x

e dx

x





 Integration by Parts:

The formula for the integration of a product of two functions is referred to as integration by parts. that is,

 

^{uv dx u vdx du vdx dx}

dx

 

   

 

   

While applying the above rule for integration by parts to the product of two functions, care should be taken to choose properly the first function, that is, the function not to be integrated.

Problem: Solve the following integrations, (i)



xe dx^x (ii) ^{ln ln}

 

^x

x dx



Solution:

(i) Given that,



xe dx^x Let, I 



xe dx^x

x dx x

x e dx e dx dx dx

 

   

 

  

x 1. x

xe e dx

 



x x

xe e c

  

where c is an integrating constant.

(ii) Given that, ^{ln ln}

 

^x

x dx



Let, ^{ln ln}

 

^x

I dx





x

Now, put 1

ln dz

z x

dx x

   dx

dz x

  Now I 



ln( )z dz 



ln( ) 1z  dz

ln( ) 1 d ln( ) 1

z dz z dz dz

dz

 

   

 

  

.

ln( )z z 1 z dz ln( )z z 1dz z

 

  



     



ln( )z z z c ln(ln ) lnx x lnx c

       

Exercise: Solve the following integrations, (i)



x²cosxdx

The definite integral defined in a bounded interval is a finite number. In this section we are going to concentrate on the definite integral and its properties.

Definite integral:

If f x( )is a continuous function defined in the interval



^{a b,}



then the definite integral with respect to x is defined as,

Fundamental Theorem of Integral Calculus:

If ^{f x}

 

be a bounded and continuous function defined in the interval

 

^{a b,} where, b >a and there exists a function ^

 

^x such that ^'

 

^{x  f x}

 

^{, then}

Definite Integration

.

     

b

a

f x dx b  a



This is called the fundamental theorem of integral calculus.

Integration as the limit of a sum:

Let, ^{f x}

 

be a bounded and continuous function defined in the interval

 

^{a b,} where a, b are finite quantities and ba.

If the interval

 

^{a b,} be divided into n equal sub-intervals, each of length h, by the points

 

, 2 , 1

a h a  h a n h so that nh b athen the area enclosed by ^{f x}

 

is defined as

         

0

lim 2 1

h hf a hf a h hf a h hf a n h

          

¹

 

0 0

lim ,

n

h r

h f a rh where nh b a



 





  

which is also defined as the definite integral of ^{f x}

 

with respect to x between the limits a and b , and is denoted by the symbol, ^b

 

a

f x dx



where, a is called the lower limit and b is called the upper limit.

Therefore,

 

¹

 

0 0

lim ,

b n

h r

a

f x dx h f a rh where nh b a



 





  



General Properties of the Definite Integral:

1.



a^b

 

^

  

b

a f t dt dx

x f

2.



a^b

 

^{ }

  

a

b f x dx dx

x f

3.



a^b

 

^

  

^

  

b c c

a f x dx f x dx

dx x

f where c is any point in the interval



a,b



.

4.



0^{a f}

 

^{x dx }



0^{a f}



^ax



^dx

.

Example: Solve the following problems,

(i)

1 2 3

(6x 5x 2)dx



 



(ii)

log 2

0 1

x x

e dx

e



(iii)

3

0

(2sin 5cos )d



   



(iv)

2

1

( ln )x x dx



Solution:

(i) Given that,

1 2 3

(6x 5x 2)dx



 



1 1

2 2

3 3

(6x 5x 2)dx (6x 5x 2)dx

 

    

 

3 2

1

[6 5 2 ]3

3 2

x x

x _

  

2 2

3 1 3 ( 3)

(2 1 5 2 1) (2 ( 3) 5 2 ( 3))

2 2

            

5 135

(2 2) ( 54 6)

2 2

       3 15

2 2

 

1 2 3

(6 5 2) 12 x x dx 2







   

(ii) Given that,

log 2

0 1

x x

e dx

e



log 2

log 2 0 0

log(1 ) 1

x

x x

e dx e

e   





[

^{( )} ln ( )

( )

f x dx f x f x

 

 ]

log 2 0

{log(1 e ) log(1 e )}

   

{log(1 2) log(1 1)}

   

(log 3 log 2)

 

(log 3 log 2)

 

log 2

0

log3

1 2

x x

e dx

 e 





(iii) Given that,

3

0

(2sin 5cos )d



   



Note: We should be careful about the use of the mathematical operators log and ln. The operator ln is used specially for expression and log is used for numerical value. But in case of expression we have to consider log as ln.

.

3

3 0 0

(2sin 5cos )d [ 2 cos 5sin ]



       



( 2 cos 5sin ) ( 2 cos 0 5sin 0)

3 3

 

     

1 3 5 3

( 2 5 ) ( 2 1 0) 1 2

2 2 2

          

3

0

(2sin 5cos ) 1 5 3

d 2



  





  

(iv) Given that,

2

1

( ln )x x dx



2 2 2

1 1 1

( ln ) ln ( d ln )

x x dx x x dx x x dx dx

  dx

   

2 2

2

[ln ]1

2 4

x x

 x 

4 4 1 1

(ln 2 ) (ln1 )

2 4 2 4

     

1 1

2 ln 2 1 (8ln 2 3)

4 4

    

2

1

( ln ) 1(8ln 2 3) x x dx 4





 

Finding Definite Integration By Using Gamma Beta Function:

Beta Function or First Eulerian Integral: A function of the form,

 

1 1 1 0

1 ⁿ ; , 0

xm^ x ^ dx m n



is called Beta function or first Eulerian integral and it is denoted by, ^



^{m n,}



^.

 

^{1 1}

 

¹

0

. , , ^m 1 ⁿ ; , 0

i e  m n 



x ^ x ^ dx m n ^.

.

Gamma Function or Second Eulerian Integral: A function of the form,

1 0

; 0

x n

e x dx n

   



is called Gamma function or second Eulerian integral and it is denoted by, ( )n .

¹

0

. , ( ) ^{x n} ; 0

i e n e x dx n

  

 



 ^.

Properties of Beta and Gamma functions: The properties are given below:

1. (1) 1

2.    (n 1) n ( ) ;n n0 3.



^,



^{( ) ( )}

( )

m n

m n m n

  ^{ }

  4. (1 )

2 

  .

5.

2

0

1 1

2 2

sin cos

2 2 2

p q

p q

x x dx

p q

  

  



^.

Problem-01: Evaluate

2 7 0

cos x dx





2

0 7

0

sin xcos x dx







0 1 7 1

2 2

0 7 2

2 2

 

  

1 8 2 2 2 9

2



1 4 2 2 9

2



1 3 2 1 2

7 5 3 1 1 2 . . . .

2 2 2 2 2

  

.

3.2.1 7 5 3 1

2 . . .

2 2 2 2

 16

35 (Ans.)

Problem-02: Evaluate

2 6 0

sin x dx





Solution: Solution: Let,

2 6 0

sin

I x dx







2

6 0

0

sin xcos x dx







6 1 0 1

2 2

6 0 2

2 2

 

  

7 1 2 2 2 8

2



7 1

2 2

2 4



5 3 1 1 1 . . . . 2 2 2 2 2

2. 3.2.1



5 3 1 . . . . 2 2 2

2. 3.2.1

   ^5.

32

  (Ans.)

Problem-03: Evaluate

2

4 3

0

sin xcos x dx





Solution: Let,

2

4 3

0

sin cos

I x x dx







4 1 3 1

2 2

4 3 2

2 2

 

  

.

5 4 2 2 2 9

2



3 1 1 2 2 2. .

7 5 3 1 1 2. . . .

2 2 2 2 2

 ²

35 (Ans.)

Exercise: Solve this integration,

2

6 8

0

sin xcos xdx





^.

Double and Triple Integration Problem: Solve the following Multiple Integrations,

(𝑖) ∫ ∫₁³ ₂⁴40 − 2𝑥𝑦𝑑𝑦𝑑𝑥

 

2

22

1

(ii) 4 2

y

y

x y dxdy

 

(𝑖𝑖𝑖) ∫ ∫ ∫ 3𝑥𝑦₁⁴ ₋₁³ ₀^{2 3}𝑧²𝑑𝑧𝑑𝑥𝑑𝑦 Answer:

(𝑖) Given that the double integration is, ∫ ∫ 40 − 2𝑥𝑦₁³ ₂⁴ 𝑑𝑦𝑑𝑥

= ∫ [∫ 40 − 2𝑥𝑦

4 2

𝑑𝑦]

3 1

𝑑𝑥

= ∫ [40𝑦 − 2𝑥.𝑦² 2]

2 3 4

1

𝑑𝑥

= ∫ [(40 ∙ 4 − 𝑥 ∙ 4^{3 2}) − (40 ∙ 2 − 𝑥

1 ∙ 2²)] 𝑑𝑥

= ∫ (160 − 16𝑥 − 80 + 4𝑥)³

1

𝑑𝑥

= ∫ (80 − 12𝑥)³

1

𝑑𝑥

= [80𝑥 −12𝑥² 2 ]

1 3

= (80 ∙ 3 − 6 ∙ 3²) − (80 ∙ 1 − 6 ∙ 1²)

= 112

(𝑖𝑖𝑖) Given that the triple integration is, ∫ ∫ ∫ 3𝑥𝑦₁⁴ ₋₁³ ₀^{2 3}𝑧²𝑑𝑧𝑑𝑥𝑑𝑦

= 3 ∫ ∫ [𝑥𝑦³𝑧³ 3]

0 3 2

−1

𝑑𝑥

4 1

𝑑𝑦

= ∫ ∫ 8𝑥𝑦³

3

−1 4 1

𝑑𝑥𝑑𝑦

(ii) Given that the double integration is,

 

2

2

2 2

1

2 2 2

1

4 2

4. 2

2

y

y

y

y

x y dxdy

x yx dy



 

   

 

 



2

2 2 2 1

2 2

y

y

x yx dy

 





  

2

2 2 2 1

2

y

y

x yx dy

 





  

²



^{2 2 4 3}



1

2 4y 2y y y dy





  

²



^{2 4 3}



1

2 2y y y dy





 

3 5 4 2

1

2 2. 3 5 4

y y y

 

    

 

8 32 16 2 1 1 2 2.3 5 4 3 5 4

 

       

16 32 16 2 1 1

2 3 5 4 3 5 4

 

       

 

133 133

2 60 30

   (Answer)

.

= 8 ∫ [𝑥² 2 𝑦³]

−1 4 3

1

𝑑𝑦

= 4 ∫ 9𝑦³− 𝑦³

4

1 𝑑𝑦

= 4 ∫ 8𝑦³

4 1

𝑑𝑦

= 4 [8𝑦³ 4 ]

1 4

= 8. 4⁴− 8.1 = 2040

Exercise: (𝑖) ∫ ∫ ∫ 𝑥𝑦𝑧₁² ₋₁³ ₂¹ 𝑑𝑧 𝑑𝑥 𝑑𝑦