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Lesson 1
Antiderivatives and the
Power Formula
Objectives:
At the end of the lesson, the student must be able to
β€’ Define antidifferentiation or integration
β€’ Be familiar with the properties of indefinite integrals
β€’ Perform basic integration by applying the power
formula
β€’ Perform integration using simple substitution
Antiderivatives
A function 𝐹 is called an antiderivative (or integral) of the function 𝑓 on a given interval
if 𝐹! π‘₯ = 𝑓 π‘₯ for every value of π‘₯ in the interval.
For example, the function 𝐹 π‘₯ =
"
#
π‘₯# is an antiderivative of 𝑓 π‘₯ = π‘₯$ on the interval
βˆ’βˆž, +∞ since for each π‘₯ in the given interval 𝐹! π‘₯ =
%
%&
"
#
π‘₯# = π‘₯$, which is equal
to 𝑓 π‘₯ .
However, 𝐹 π‘₯ =
"
#
π‘₯# is not the only antiderivative of 𝑓 on the given interval. If we add
any constant 𝐢 to
"
#
π‘₯# , the function 𝐺! π‘₯ =
%
%&
"
#
π‘₯# + 𝐢 = π‘₯$ = 𝑓 π‘₯ .
In general, once any single antiderivative is known, the other antiderivatives can
be obtained by adding constants to the known derivatives. Thus,
"
#
π‘₯#,
"
#
π‘₯# + 3,
"
#
π‘₯# + πœ‹,
"
#
π‘₯# + 5 ,
"
#
π‘₯# βˆ’ 2 are all antiderivatives of 𝑓 π‘₯ = π‘₯$.
Theorem:
If 𝐹 π‘₯ is any antiderivative of 𝑓 π‘₯ on an open interval, then for any
constant 𝐢, the function 𝐹 π‘₯ + 𝐢 is also an antiderivative on that interval.
Moreover, each antiderivative of 𝑓 π‘₯ on the interval can be expressed in the
form 𝐹 π‘₯ + 𝐢 by choosing the constant 𝐢 appropriately.
Definition: The Indefinite Integral
The process of finding antiderivatives is called antidifferentiation or integration.
Thus, if
%
%&
𝐹 π‘₯ = 𝑓 π‘₯ , then integrating (or antidifferentiating) the function 𝑓 π‘₯
produces an antiderivative of the form 𝐹 π‘₯ + 𝐢. To emphasize this process, we use
the following notation,
∫ 𝑓 π‘₯ 𝑑π‘₯ = 𝐹 π‘₯ + 𝐢
where,
∫ 𝑓 π‘₯ 𝑑π‘₯ is the antiderivative
∫ is the integral sign
𝑓 π‘₯ is the integrand
𝐢 is the constant of integration
𝑑π‘₯ indicates that π‘₯ is the variable of integration
Some of the properties of indefinite integral and basic integration formulas, which need
no proof from the fact that these properties are also known properties of
differentiation, are listed below.
Properties of Indefinite Integral and Basic Integration Formula:
i. ∫ 𝑑π‘₯ = π‘₯ + 𝐢 ( Definition of an integral)
ii. ∫ 𝑐 𝑓 π‘₯ 𝑑π‘₯ = 𝑐 ∫ 𝑓 π‘₯ 𝑑π‘₯ = 𝑐 𝐹 π‘₯ + 𝐢
iii. ∫ 𝑓" π‘₯ Β± 𝑓' π‘₯ Β± β‹― Β± 𝑓( π‘₯ 𝑑π‘₯ = ∫ 𝑓" π‘₯ 𝑑π‘₯ Β± ∫ 𝑓' π‘₯ 𝑑π‘₯ Β± β‹― Β± ∫ 𝑓( π‘₯ 𝑑π‘₯
iv. ∫ π‘₯( 𝑑π‘₯ =
&!"#
()"
+ 𝐢 , 𝑛 β‰  βˆ’1 ( The Power Formula )
The General Power Formula
In evaluating ∫ 𝑓 𝑔 π‘₯ 𝑔! π‘₯ 𝑑π‘₯ , it will be more convenient to let 𝑒 = 𝑔 π‘₯ and write the
differential form 𝑑𝑒 = 𝑔′ π‘₯ 𝑑π‘₯ . Thus,
∫ 𝑓 𝑒 𝑑𝑒 = 𝐹 𝑒 + 𝐢
The generalized power formula therefore is:
, 𝑓 𝑒 "𝑑 𝑓 𝑒 =
𝑓 𝑒 "#$
𝑛 + 1
+ 𝐢 𝑛 β‰  βˆ’1
or
7 𝑒(𝑑𝑒 =
𝑒()"
𝑛 + 1
+ 𝐢 𝑛 β‰  βˆ’1
The method of u-substitution may be applied in evaluating an integral with the substitution 𝑒 =
𝑔 π‘₯ and 𝑑𝑒 = 𝑔! π‘₯ 𝑑π‘₯
Examples:
1. Evaluate ∫ 6π‘₯' βˆ’ 4π‘₯ + 5 𝑑π‘₯
Solution:
∫ 6π‘₯' βˆ’ 4π‘₯ + 5 𝑑π‘₯ = 6 ∫ π‘₯' 𝑑π‘₯ βˆ’ 4 ∫ π‘₯ 𝑑π‘₯ + 5 ∫ 𝑑π‘₯
= 6 ;
&$
$
βˆ’ 4 ;
&%
'
+ 5π‘₯ + 𝐢
= 2π‘₯$ βˆ’ 2π‘₯' + 5π‘₯ + 𝐢
2. Evaluate ∫ 4 𝑑 +
'
*
𝑑𝑑
Solution:
∫ 4 𝑑 +
'
*
𝑑𝑑 = 4 ∫ 𝑑
#
% 𝑑𝑑 + 2 ∫
"
*
#
%
𝑑𝑑 = 4 ∫ 𝑑
#
% 𝑑𝑑 + 2 ∫ 𝑑+
#
% 𝑑𝑑
= 4 ;
*
$
%
$
%
+ 2 ;
*
#
%
#
%
+ 𝐢 =
,
$
𝑑
$
% + 4𝑑
#
% + 𝐢
3. Evaluate ∫ 3π‘₯ + 1 ' 𝑑π‘₯
Solution 1:
∫ 3π‘₯ + 1 ' 𝑑π‘₯ = ∫ 9π‘₯' + 6π‘₯ + 1 𝑑π‘₯
= 9 ∫ π‘₯' 𝑑π‘₯ + 6 ∫ π‘₯ 𝑑π‘₯ + ∫ 𝑑π‘₯
= 9 ;
&$
$
+ 6 ;
&%
'
+ π‘₯ + 𝐢
= 3 π‘₯$ + 3π‘₯' + π‘₯ + 𝐢
Solution 2:
∫ 3π‘₯ + 1 ' 𝑑π‘₯ =
"
$
∫ 3π‘₯ + 1 ' ; 3 𝑑π‘₯
=
"
$
;
$&)" $
$
+ 𝐢
=
"
-
3π‘₯ + 1 $ + 𝐢
Solution 3: (by u-substitution)
∫ 3π‘₯ + 1 ' 𝑑π‘₯ =
"
$
∫ 𝑒' 𝑑𝑒
=
"
$
;
.$
$
+ 𝐢
=
"
-
3π‘₯ + 1 $ + 𝐢
4. Evaluate ∫
&$ %&
'&&+/ '
Solution:
∫
&$ %&
'&&+/ ' = ∫ 2π‘₯# βˆ’ 5 +0 π‘₯$𝑑π‘₯
=
"
,
∫ 𝑒+0 𝑑𝑒
=
"
,
;
.()
+1
+ 𝐢 = βˆ’
"
#, '&&+/ ) + 𝐢
5. Evaluate ∫
')$ 234 5 & %5
678%5
Solution:
∫
')$ 234 5 & %5
678%5
= ∫ 2 + 3 tan 𝑦 # 𝑠𝑒𝑐'𝑦 𝑑𝑦
=
"
$
∫ 𝑒# 𝑑𝑒
=
"
$
;
.*
/
+ 𝐢 =
"
"/
2 + 3 tan 𝑦 / + 𝐢
6. Evaluate ∫ cot π‘₯ ln sin π‘₯ 𝑑π‘₯
Solution:
∫ cot π‘₯ ln sin π‘₯ 𝑑π‘₯ = ∫ 𝑒 𝑑𝑒
=
.%
'
+ 𝐢
=
9(% :;4 &
'
+ 𝐢
7. Evaluate ∫
#&+" %&
'&)$ * &+' *
Solution:
∫
#&+" %&
'&)$ * &+' * = ∫
#&+" %&
'&)$ &+' *
= ∫
#&+" %&
'&%+&+1 *
= ∫ 2π‘₯' βˆ’ π‘₯ βˆ’ 6 +/ 4π‘₯ βˆ’ 1 𝑑π‘₯
= ∫ 𝑒+/ 𝑑𝑒
=
.(&
+#
+ 𝐢
=
'&%+&+1
(&
+#
+ 𝐢
= βˆ’
"
# '&%+&+1 & + 𝐢
Exercises:
Evaluate the following integrals:
1. ∫ 𝑑 βˆ’
$
𝑑' +
*
𝑑$ 𝑑𝑑 2. ∫
234 '& %&
$)<4 =>: '& )
Exercises:
3. ∫
")$?(%+ ,
%&
?%+ 4. ∫ cot 3π‘₯ 𝑐𝑠𝑐$3π‘₯ 𝑑π‘₯
Exercises:
5. 7
2π‘₯ + 1 𝑑π‘₯
π‘₯ + 4 1 π‘₯ βˆ’ 3 1 6. 7 𝑠𝑖𝑛# 𝑒$& cos 𝑒$& 𝑒$& 𝑑π‘₯

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Lesson 1 Antiderivatives and the Power Formula.pdf

  • 1. Lesson 1 Antiderivatives and the Power Formula
  • 2. Objectives: At the end of the lesson, the student must be able to β€’ Define antidifferentiation or integration β€’ Be familiar with the properties of indefinite integrals β€’ Perform basic integration by applying the power formula β€’ Perform integration using simple substitution
  • 3. Antiderivatives A function 𝐹 is called an antiderivative (or integral) of the function 𝑓 on a given interval if 𝐹! π‘₯ = 𝑓 π‘₯ for every value of π‘₯ in the interval. For example, the function 𝐹 π‘₯ = " # π‘₯# is an antiderivative of 𝑓 π‘₯ = π‘₯$ on the interval βˆ’βˆž, +∞ since for each π‘₯ in the given interval 𝐹! π‘₯ = % %& " # π‘₯# = π‘₯$, which is equal to 𝑓 π‘₯ . However, 𝐹 π‘₯ = " # π‘₯# is not the only antiderivative of 𝑓 on the given interval. If we add any constant 𝐢 to " # π‘₯# , the function 𝐺! π‘₯ = % %& " # π‘₯# + 𝐢 = π‘₯$ = 𝑓 π‘₯ .
  • 4. In general, once any single antiderivative is known, the other antiderivatives can be obtained by adding constants to the known derivatives. Thus, " # π‘₯#, " # π‘₯# + 3, " # π‘₯# + πœ‹, " # π‘₯# + 5 , " # π‘₯# βˆ’ 2 are all antiderivatives of 𝑓 π‘₯ = π‘₯$. Theorem: If 𝐹 π‘₯ is any antiderivative of 𝑓 π‘₯ on an open interval, then for any constant 𝐢, the function 𝐹 π‘₯ + 𝐢 is also an antiderivative on that interval. Moreover, each antiderivative of 𝑓 π‘₯ on the interval can be expressed in the form 𝐹 π‘₯ + 𝐢 by choosing the constant 𝐢 appropriately.
  • 5. Definition: The Indefinite Integral The process of finding antiderivatives is called antidifferentiation or integration. Thus, if % %& 𝐹 π‘₯ = 𝑓 π‘₯ , then integrating (or antidifferentiating) the function 𝑓 π‘₯ produces an antiderivative of the form 𝐹 π‘₯ + 𝐢. To emphasize this process, we use the following notation, ∫ 𝑓 π‘₯ 𝑑π‘₯ = 𝐹 π‘₯ + 𝐢 where, ∫ 𝑓 π‘₯ 𝑑π‘₯ is the antiderivative ∫ is the integral sign 𝑓 π‘₯ is the integrand 𝐢 is the constant of integration 𝑑π‘₯ indicates that π‘₯ is the variable of integration
  • 6. Some of the properties of indefinite integral and basic integration formulas, which need no proof from the fact that these properties are also known properties of differentiation, are listed below. Properties of Indefinite Integral and Basic Integration Formula: i. ∫ 𝑑π‘₯ = π‘₯ + 𝐢 ( Definition of an integral) ii. ∫ 𝑐 𝑓 π‘₯ 𝑑π‘₯ = 𝑐 ∫ 𝑓 π‘₯ 𝑑π‘₯ = 𝑐 𝐹 π‘₯ + 𝐢 iii. ∫ 𝑓" π‘₯ Β± 𝑓' π‘₯ Β± β‹― Β± 𝑓( π‘₯ 𝑑π‘₯ = ∫ 𝑓" π‘₯ 𝑑π‘₯ Β± ∫ 𝑓' π‘₯ 𝑑π‘₯ Β± β‹― Β± ∫ 𝑓( π‘₯ 𝑑π‘₯ iv. ∫ π‘₯( 𝑑π‘₯ = &!"# ()" + 𝐢 , 𝑛 β‰  βˆ’1 ( The Power Formula )
  • 7. The General Power Formula In evaluating ∫ 𝑓 𝑔 π‘₯ 𝑔! π‘₯ 𝑑π‘₯ , it will be more convenient to let 𝑒 = 𝑔 π‘₯ and write the differential form 𝑑𝑒 = 𝑔′ π‘₯ 𝑑π‘₯ . Thus, ∫ 𝑓 𝑒 𝑑𝑒 = 𝐹 𝑒 + 𝐢 The generalized power formula therefore is: , 𝑓 𝑒 "𝑑 𝑓 𝑒 = 𝑓 𝑒 "#$ 𝑛 + 1 + 𝐢 𝑛 β‰  βˆ’1 or 7 𝑒(𝑑𝑒 = 𝑒()" 𝑛 + 1 + 𝐢 𝑛 β‰  βˆ’1 The method of u-substitution may be applied in evaluating an integral with the substitution 𝑒 = 𝑔 π‘₯ and 𝑑𝑒 = 𝑔! π‘₯ 𝑑π‘₯
  • 8. Examples: 1. Evaluate ∫ 6π‘₯' βˆ’ 4π‘₯ + 5 𝑑π‘₯ Solution: ∫ 6π‘₯' βˆ’ 4π‘₯ + 5 𝑑π‘₯ = 6 ∫ π‘₯' 𝑑π‘₯ βˆ’ 4 ∫ π‘₯ 𝑑π‘₯ + 5 ∫ 𝑑π‘₯ = 6 ; &$ $ βˆ’ 4 ; &% ' + 5π‘₯ + 𝐢 = 2π‘₯$ βˆ’ 2π‘₯' + 5π‘₯ + 𝐢 2. Evaluate ∫ 4 𝑑 + ' * 𝑑𝑑 Solution: ∫ 4 𝑑 + ' * 𝑑𝑑 = 4 ∫ 𝑑 # % 𝑑𝑑 + 2 ∫ " * # % 𝑑𝑑 = 4 ∫ 𝑑 # % 𝑑𝑑 + 2 ∫ 𝑑+ # % 𝑑𝑑 = 4 ; * $ % $ % + 2 ; * # % # % + 𝐢 = , $ 𝑑 $ % + 4𝑑 # % + 𝐢
  • 9. 3. Evaluate ∫ 3π‘₯ + 1 ' 𝑑π‘₯ Solution 1: ∫ 3π‘₯ + 1 ' 𝑑π‘₯ = ∫ 9π‘₯' + 6π‘₯ + 1 𝑑π‘₯ = 9 ∫ π‘₯' 𝑑π‘₯ + 6 ∫ π‘₯ 𝑑π‘₯ + ∫ 𝑑π‘₯ = 9 ; &$ $ + 6 ; &% ' + π‘₯ + 𝐢 = 3 π‘₯$ + 3π‘₯' + π‘₯ + 𝐢 Solution 2: ∫ 3π‘₯ + 1 ' 𝑑π‘₯ = " $ ∫ 3π‘₯ + 1 ' ; 3 𝑑π‘₯ = " $ ; $&)" $ $ + 𝐢 = " - 3π‘₯ + 1 $ + 𝐢
  • 10. Solution 3: (by u-substitution) ∫ 3π‘₯ + 1 ' 𝑑π‘₯ = " $ ∫ 𝑒' 𝑑𝑒 = " $ ; .$ $ + 𝐢 = " - 3π‘₯ + 1 $ + 𝐢 4. Evaluate ∫ &$ %& '&&+/ ' Solution: ∫ &$ %& '&&+/ ' = ∫ 2π‘₯# βˆ’ 5 +0 π‘₯$𝑑π‘₯ = " , ∫ 𝑒+0 𝑑𝑒 = " , ; .() +1 + 𝐢 = βˆ’ " #, '&&+/ ) + 𝐢
  • 11. 5. Evaluate ∫ ')$ 234 5 & %5 678%5 Solution: ∫ ')$ 234 5 & %5 678%5 = ∫ 2 + 3 tan 𝑦 # 𝑠𝑒𝑐'𝑦 𝑑𝑦 = " $ ∫ 𝑒# 𝑑𝑒 = " $ ; .* / + 𝐢 = " "/ 2 + 3 tan 𝑦 / + 𝐢 6. Evaluate ∫ cot π‘₯ ln sin π‘₯ 𝑑π‘₯ Solution: ∫ cot π‘₯ ln sin π‘₯ 𝑑π‘₯ = ∫ 𝑒 𝑑𝑒 = .% ' + 𝐢 = 9(% :;4 & ' + 𝐢
  • 12. 7. Evaluate ∫ #&+" %& '&)$ * &+' * Solution: ∫ #&+" %& '&)$ * &+' * = ∫ #&+" %& '&)$ &+' * = ∫ #&+" %& '&%+&+1 * = ∫ 2π‘₯' βˆ’ π‘₯ βˆ’ 6 +/ 4π‘₯ βˆ’ 1 𝑑π‘₯ = ∫ 𝑒+/ 𝑑𝑒 = .(& +# + 𝐢 = '&%+&+1 (& +# + 𝐢 = βˆ’ " # '&%+&+1 & + 𝐢
  • 13. Exercises: Evaluate the following integrals: 1. ∫ 𝑑 βˆ’ $ 𝑑' + * 𝑑$ 𝑑𝑑 2. ∫ 234 '& %& $)<4 =>: '& )
  • 14. Exercises: 3. ∫ ")$?(%+ , %& ?%+ 4. ∫ cot 3π‘₯ 𝑐𝑠𝑐$3π‘₯ 𝑑π‘₯
  • 15. Exercises: 5. 7 2π‘₯ + 1 𝑑π‘₯ π‘₯ + 4 1 π‘₯ βˆ’ 3 1 6. 7 𝑠𝑖𝑛# 𝑒$& cos 𝑒$& 𝑒$& 𝑑π‘₯