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Integration (Basic Integration)

This document discusses techniques for integrating algebraic expressions and using substitution to evaluate indefinite and definite integrals. Some key points are: 1) Indefinite integrals of algebraic expressions can be evaluated by converting them to the standard form of ax^n and using the formula ∫ax^n dx = a/(n+1)x^(n+1) + c. 2) The substitution method allows one to make u-substitutions to simplify integrals and convert them to forms that can be integrated. 3) Definite integrals are evaluated by substituting the limits of integration and finding the antiderivative evaluated between those limits.

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Fazlina Mustafa
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99 views6 pages

Integration (Basic Integration)

This document discusses techniques for integrating algebraic expressions and using substitution to evaluate indefinite and definite integrals. Some key points are: 1) Indefinite integrals of algebraic expressions can be evaluated by converting them to the standard form of ax^n and using the formula ∫ax^n dx = a/(n+1)x^(n+1) + c. 2) The substitution method allows one to make u-substitutions to simplify integrals and convert them to forms that can be integrated. 3) Definite integrals are evaluated by substituting the limits of integration and finding the antiderivative evaluated between those limits.

Uploaded by

Fazlina Mustafa
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOC, PDF, TXT or read online on Scribd
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Integration

Integration of Simple Polinomial


Exp.1
5 dx = 5x
0
dx
= 5 x
0+1
+ c

0+1
= 5x + c
Convert to the form of ax
n
before
integrate
ax
n
dx = ax
n+1
+ c

n+1
(a and c are constants)
Exp.
!1!! dx = "x
#
dx
x

= x
#+1
+ c
(
#+1)

= #" x
#1
+ c
= #1 + c
x
Convert to the form of ax
n
before
integrate
ax
n
dx = ax
n+1
+ c

n+1
Try it yourself
a)#$dx= Convert to the form of ax
n
before
integrate
ax
n
dx = ax
n+1
+ c

n+1
b) #10x
5
dx= ax
n
dx = ax
n+1
+ c

n+1
c) 5! dx =
$x
%
Convert to the form of ax
n
before
integrate
ax
n
dx = ax
n+1
+ c

n+1
d) % x
%
dx =

e) #$ x
"
dx =

f) 5!



dx =
$ x
%
g) 1

dx =
( x)

h) % & x

dx =

i) x
1'%
dx =

() #)

dx =
5x
5
*) #+ dx =
&x
,) (& x + x)

dx = m) ( x

- x +% )

dx =
A.Integrals of algebraic expressions
a) (%x - )(x+1) dx = (%x

+ x - )dx
= %x
%


+ x

# x + c
%


= x
%
+ "x

# x + c
expand' simp,if. expression to the
,o/est form
ax
n
dx = ax
n+1
+ c

n+1
(a and c are constants)
b) x

+ %x dx = ( x + % )dx
x
=

x

+ %x + c



expand' simp,if. expression to the
,o/est form
ax
n
dx = ax
n+1
+ c

n+1
Try it yourself
c) (x

- 1)( #x) dx= d) x

# % dx =
x
5
e) x( x # ") dx = f) x

( x

# %) dx =

g) + # x dx =
x
%
h) 0 - x

dx =
% + x
B.Integration by Substitution
a) (4x #%)
%
dx
= 1
%
(1'4 d1)
= 2 1
%
d1
= 2 ( 1
+
) + c
+
= 1 (+x - %)
+
+ c
1)
3et u = 4x -3
4hen du/dx = 4
(ax + b)
n
dx = (ax + b)
n+1
+ c
a(n+1)
(a5 b and c are constants)
ax
n
dx = ax
n+1
+ c

n+1
s1bstit1te bac* u = 4x -3
b) 1 dx =
(%x + )
+
c) % dx =
($ - %x )

d) (6 - +x)
)
dx =

e) x #

dx

f) 7 ( - )x)
%
dx g) #% dx =
(x + 10)
6
C.Definite Integrals
a)

#$dx=

#$x
0

1 1
= 8 #$x 9


1
= 8#$()9 - 8#$(1)9
= #$
Convert to the form of ax
n
before
integrate
ax
n
dx = ax
n+1
+ c

n+1
:1bstit1te 8;pper ,imit9#83o/er ,imit9
b)
%
#10x
5
dx =
0
c)
0
(#x)(x + %) dx

#
d)
%
5! dx =

#1
$x
%
e)

(& x + x)

dx =

#1
d)
1
% x
%
dx =

#

e)

#$ x
"
dx =

1

f)
%
5!



dx =

0
$ x
%
g)
%
1

dx =

( x)

h) %

& x

dx =

#1
i)
%
x
1'%
dx =

1

()

#)

dx =

1
5x
5
*)
%
#+ dx =

1
&x
,)
#
x #

dx

#%

m)
+
( x

- x +% )

dx =

0

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