calculus - iiw.kuleuven.be · calculus 3 1 relations, functions, mappings, 1-1 mappings in...
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Calculus
Dr. Caroline Danneels
1 Relations, functions, mappings, 1-1 mappings ..................................................................3
2 Expanding IR ........................................................................................................................4
3 Continuity of a function in IR .............................................................................................5
3.1 Continuity at a point ................................................................................................................... 5
3.2 Right- and left-continuity ........................................................................................................... 6
4 Limits .....................................................................................................................................7
4.1 Limits in a real number .............................................................................................................. 7
4.2 Right-hand and left-hand limits ................................................................................................. 9
4.3 Limits at infinity ........................................................................................................................ 9
4.4 Infinite limits .............................................................................................................................. 9
4.5 Infinite limits at infinity ........................................................................................................... 10
4.6 Elementary rules to calculate limits ......................................................................................... 10
4.7 Indeterminate cases .................................................................................................................. 11
4.8 Exercises .................................................................................................................................. 14
5 Differentiation .....................................................................................................................17
5.1 The derivative in a point .......................................................................................................... 17
5.2 Geometrical interpretation of the derivative ............................................................................ 17
5.3 Left-hand and right-hand derivative in a point ........................................................................ 18
5.4 A vertical tangent ..................................................................................................................... 19
5.5 Differentiation rules ................................................................................................................. 19
5.6 Exercises .................................................................................................................................. 21
6 Indefinite integrals .............................................................................................................24
6.1 Antiderivative functions ........................................................................................................... 24
6.2 The indefinite integral .............................................................................................................. 24
6.3 Properties ................................................................................................................................. 24
6.4 Basic integrals .......................................................................................................................... 25
6.5 Integration by substitution ....................................................................................................... 25
6.6 Exercises .................................................................................................................................. 26
6.7 Integration by parts .................................................................................................................. 29
6.8 Exercises .................................................................................................................................. 29
7 Definite integrals ................................................................................................................30
7.1 The fundamental theorem of calculus ...................................................................................... 30
7.2 Properties of the definite integral ............................................................................................. 30
7.3 The substitution method ........................................................................................................... 31
7.4 Partial integration ..................................................................................................................... 31
7.5 Exercices .................................................................................................................................. 32
Calculus 3
1 Relations, functions, mappings, 1-1 mappings
In mathematics, a relation from A to B is a collection of ordered pairs (x,y) where x A∈ and y B∈ , in other words it is a subset of A x B.
The domain of the relation R is the set of all the first numbers of the ordered pairs. In other words: dom R { }| ( , )x x y R= ∈ .
The range of the relation R is the set of the second numbers in each pair. In other words: range R { }| ( , )y x y R= ∈ .
If one switches the order of the pairs of a relation, then one gets the inverse relation from B to A.
A function is a special type of relation where each x-value has one and only one y-value, so no
x-value can be repeated. All functions are relations but not all relations are functions.
The x-number is called the independent variable, and the y-number is called the dependent
variable because its value depends on the x-value chosen.
A function where for every element x of A there is exactly one f(x) in B is called a mapping
from A to B. Then dom f = A.
A mapping from A to B is called a 1-1 mapping when the inverse is also a mapping from B to A.
Calculus 4
2 Expanding IR
Given the set of real numbers IR, we add two elements −∞ en +∞ , and call the new set the
expanded set of real numbers.
Notatie: { },IR IR= ∪ −∞ +∞
It is clear that :x IR x∀ ∈ −∞ < < +∞ .
Properties:
0
0
1. : ( ) ( )
( ) ( )
: ( ) ( )
( ) ( )
: ( ) ( )
( ) ( )
x IR x x
x x
x IR x x
x x
x IR x x
x x
+
−
∀ ∈ + −∞ = −∞ = −∞ ++ +∞ = +∞ = +∞ +
∀ ∈ ⋅ +∞ = +∞ = +∞ ⋅⋅ −∞ = −∞ = −∞ ⋅
∀ ∈ ⋅ +∞ = −∞ = +∞ ⋅⋅ −∞ = +∞ = −∞ ⋅
2. ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
if n is odd
1 10 0
n
n
+∞ + +∞ = +∞−∞ + −∞ = −∞−∞ ⋅ +∞ = −∞ = +∞ ⋅ −∞+∞ ⋅ +∞ = +∞ = −∞ ⋅ −∞
+∞ = +∞
−∞ = −∞
= =+∞ −∞
So, the following expressions have no meaning, they are indefinite:
+∞( ) + −∞( ) ; −∞( ) + +∞( ) ; 0.+∞ ; + ∞. 0 ; 0.−∞ ; −∞.0 ;+∞+∞
;+∞−∞
;−∞+ ∞
;−∞−∞
Calculus 5
3 Continuity of a function in IR
3.1 Continuity at a point
Definition:
a function f: IR → IR is continuous in a point a if the graph of the function does not show a
jump in a.
If the function is not continuous in a, it is defined as discontinuous in a.
Examples
f is continuous in a f is discontinuous in a
f is not left-continuous in a
f is not right-continuous in a
f is discontinuous in a f is discontinuous in a
f is not left-continuous in a f is right-continuous in a
f is not right-continuous in a
f is discontinuous in a
f is left-continuous in a
f(a)
a
b
f(a)
a
Y
X
Y
X
X
Y
X
Y
a a
c
f(a)
b
f(a)
b
X
Y
a
bf(a)
Calculus 6
It is clear that the point a must belong to the domain of the function, otherwise continuity is not
possible.
The following mathematical definition can be understood in terms of the previous intuitive
definition of a continuous function.
is continuous in 0, 0 : ( ) ( )
a domff a
x a f x f aε δ δ ε∈⇔ ∀ > ∃ > − < ⇒ − <
3.2 Right- and left-continuity
is right-continuous in 0, 0 : ( ) ( )
is left-continuous in 0, 0 : ( ) ( )
a domff a
a x a f x f a
a domff a
a x a f x f a
ε δ δ ε
ε δ δ ε
∈⇔ ∀ > ∃ > ≤ < + ⇒ − <
∈⇔ ∀ > ∃ > − < ≤ ⇒ − <
Calculus 7
4 Limits
4.1 Limits in a real number
The concept ‘limit’ allows us to examine the behavior of a function in the close neighbourhood
of a point. The function value in this point is without importance.
Assume that a is a real number which is an adherent point of dom f. This means that in each interval around a, 0] , [,a a IRε ε ε +− + ∈ , there are numbers of dom f which are different from a.
In other words, f is not necessarily defined in a, but it is defined in other points arbitrary close
to a.
We say that a function has a limit b at an input a if the values f(x) come arbitrarily close to b for
x-values which come sufficiently close to a.
Notation: lim ( )
x af x b
→=
Since lim ( )
x af x b
→= describes the behavior of the function f in the environment of the point a,
the concept lim ( )x a
f x→
is obviously meaningless when a is an isolated point of dom f.
Definition:
If f(x) is defined in an open interval around a, except possibly at a itself, then
lim ( ) 0, 0 :0 ( )x a
f x b IR x a f x bε δ δ ε→
= ∈ ⇔ ∀ > ∃ > < − < ⇒ − <
Note that the limit equals the function value in a if the function is continuous in a:
lim ( ) ( ) is continuous in x a
f x f a f a→
= ⇔
Calculus 8
Examples
f is continuous in a f is discontinuous in a
f is not left-continuous in a
f is not right-continuous in a lim ( ) ( )
x af x f a
→= lim ( )
x af x b
→=
f is discontinuous in a f is discontinuous in a
f is not left-continuous in a f is right-continuous in a
f is not right-continuous in a
lim ( ) lim ( )
lim ( ) does not exist
x a x a
x a
f x c f x b
f x> <
→ →
→
= ≠ =
⇒
lim ( ) ( ) lim ( )
lim ( ) does not exist
x a x a
x a
f x f a f x b
f x> <
→ →
→
= ≠ =
⇒
f is discontinuous in a
f is left-continuous in a
lim ( ) lim ( ) ( )
lim ( ) does not exist
x a x a
x a
f x b f x f a
f x> <
→ →
→
= ≠ =
⇒
f(a)
a
b
f(a)
a
Y
X
Y
X
X
Y
X
Y
a a
c
f(a)
b
f(a)
b
X
Y
a
bf(a)
Calculus 9
4.2 Right-hand and left-hand limits
We say the left-hand (or right-hand) limit of f(x) as x approaches a is b (or the limit of f(x) as x
approaches a from the left (or the right) is b) if the values f(x) come arbitrarily close to b for x-
values which come sufficiently close to a from the left-hand side (right-hand side).
] [
] [
lim ( ) 0, 0 : , ( )
lim ( ) 0, 0 : , ( )
x a
x a
f x b x a a f x b
f x b x a a f x b
ε δ δ ε
ε δ δ ε
>
<
→
→
= ⇔ ∀ > ∃ > ∈ + ⇒ − <
= ⇔ ∀ > ∃ > ∈ − ⇒ − <
The following notations are also commonly used:
lim ( ) , lim ( ) , ( )
x aaf x b f x b f a
− ↑= = −
4.3 Limits at infinity
Limits at infinity represent the behavior of function values f(x) if |x| ever increases. To define
limits at +∞, we demand that dom f contains at least a half line of the form [a,+∞[ . To define
limits at -∞, we demand that dom f contains at least a half line of the form ]-∞,a]
We say the limit of f(x) as x approaches (negative) infinity is b if the values f(x) come arbitrarily
close to b for arbitrarily large x-values.
lim ( ) 0, 0 : ( )
lim ( ) 0, 0 : ( )
x
x
f x b m x m f x b
f x b m x m f x b
ε ε
ε ε
→+∞
→−∞
= ⇔ ∀ > ∃ > > ⇒ − <
= ⇔ ∀ > ∃ > < − ⇒ − <
4.4 Infinite limits
We say the limit of f(x) as x approaches a is (negative) infinity, if the values f(x) become
arbitrarily large for x-values which come sufficiently close to a .
lim ( ) 0, 0 : 0 ( )
lim ( ) 0, 0 : 0 ( )
x a
x a
f x n x a f x n
f x n x a f x n
δ δ
δ δ
→
→
= +∞ ⇔ ∀ > ∃ > < − < ⇒ >
= −∞ ⇔ ∀ > ∃ > < − < ⇒ < −
Calculus 10
4.5 Infinite limits at infinity
We say the limit of f(x) as x approaches (negative) infinity is (negative) infinity, if the values
f(x) become arbitrarily large for arbitrarily large x-values.
lim ( ) 0, 0 : ( )
lim ( ) 0, 0 : ( )
lim ( ) 0, 0 : ( )
lim ( ) 0, 0 : ( )
x
x
x
x
f x n m x m f x n
f x n m x m f x n
f x n m x m f x n
f x n m x m f x n
→+∞
→+∞
→−∞
→−∞
= +∞ ⇔ ∀ > ∃ > > ⇒ >
= −∞ ⇔ ∀ > ∃ > > ⇒ < −
= +∞ ⇔ ∀ > ∃ > < − ⇒ >
= −∞ ⇔ ∀ > ∃ > < − ⇒ < −
4.6 Elementary rules to calculate limits
Given two real functions f and g for which the limit in a IR∈ exists, e.g.
1 2 1 2lim ( ) and lim ( ) with ,x a x a
f x L g x L L L IR→ →
= = ∈ then
1. [ ] 1 2lim ( ) ( ) lim ( ) lim ( )
x a x a x af x g x f x g x L L
→ → →+ = + = +
2. 1lim ( ) lim ( )x a x a
k f x k f x k L→ →
⋅ = ⋅ = ⋅
3. [ ] 1 2lim ( ) ( ) lim ( ) lim ( )x a x a x a
f x g x f x g x L L→ → →
⋅ = ⋅ = ⋅
4. 12
2
lim ( )( )lim if 0
( ) lim ( )x a
x ax a
f x Lf xL
g x g x L→
→→
= = ≠
5. 1lim ( ) lim ( )x a x a
f x f x L→ →
= =
6. ( ) ( ) ( )1lim ( ) lim ( )nn n
x a x af x f x L
→ →= =
7. 1lim ( ) lim ( )n nnx a x a
f x f x L→ →
= =
8. limx a
k k→
=
Calculus 11
4.7 Indeterminate cases
4.7.1 The indeterminate case 00
Rational function f (x)
g(x):
( ) 0lim
( ) 0x a
f x
g x→
=
means that lim ( ) ( ) 0x a
f x f a→
= = and lim ( ) ( ) 0x a
g x g a→
= = . So this means that
both f(x) and g(x) are divisible by (x – a), so it is possible to write 1( ) ( ) ( )f x x a f x= − and
1( ) ( ) ( )g x x a g x= − .
therefore 1 1
1 1
( ) ( ) ( )( ) 0lim lim lim
( ) 0 ( ) ( ) ( )x a x a x a
x a f x f xf x
g x x a g x g x→ → →
− = = = −
Example 2
22 2 2
4 ( 2)( 2) 2lim lim lim 4
( 2)( 3) 35 6x x x
x x x x
x x xx x→ → →
− − + += = = −− − −− +
Irrational form: make nominator and/or denominator rational.
Example
( )3 3
3
3
1 2 ( 1 2)( 1 2)lim lim
3 ( 3)( 1 2)
1 4lim
( 3)( 1 2)
1lim
( 1 2)
1
4
x x
x
x
x x x
x x x
x
x x
x
→ →
→
→
+ − + − + +=− − + +
+ −=
− + +
=+ +
=
Calculus 12
4.7.2 The indeterminate case ∞∞∞∞∞∞∞∞
Rational function:
1
1 1 01
1 1 0
...lim lim
...
n n nn n n
P p px xP p p
a x a x a x a a x
b x b x b x b b x
−−
−→∞ →∞−
+ + + +=
+ + + +
n
p
aif
b
if (sign of has to be determined)
= 0 if
n=p
n > p
n < p
=
∞=∞
Irrational function:
Determine the highest power of x in the nominator and denominator and simplify.
Remark that
if
if
2x = x x > 0
= -x x < 0
Example
222
¨
51 2
5 2lim lim
3 3
lim
x x
x
x xxx x
x x
x
→∞ →∞
→ +∞
+ + + + =
2
51 2x
x
+ +
3 x
¨
1
limx
x
→ −∞
=
2
51 2x
x
− + +
3 x
1
3
=
4.7.3 The indeterminate case ∞ − ∞∞ − ∞∞ − ∞∞ − ∞
Polynomial:
( )11 1 0
¨ ¨lim ... limn n n
n n nx x
a x a x a x a a x−−→ ∞ → ∞
+ + + + =
Irrational function: Multiply and divide by the conjugate term.
Calculus 13
Example
( )
( )
( ) ( )( )
2
2
2 2
2
2
2
lim 4 3 1 2
/ lim 4 3 1 2
4 3 1 2 4 3 1 2/ lim 4 3 1 2 lim
4 3 1 2
3 1lim lim
4 3 1 2
x
x
x x
x x
x x x
a x x x
x x x x x xb x x x
x x x
xx
x x x
→∞
→+∞
→−∞ →−∞
→−∞ →−∞
+ − +
+ − + = +∞
+ − + + − −+ − + =
+ − −
−= =+ − −
13
x
x
−
2
3
43 14 2
x x
= −
− + − −
Calculus 14
4.8 Exercises
1. 2
4
16lim
3 12x
x
x→
−−
8
3
2. 2
23
7 12lim
4 3x
x x
x x→
− +− +
1
2−
3. 2
31
1lim
1x
x
x→
−−
2
3
4. 4 4
3 3limx a
x a
a x→
−−
4
a3
−
5. 3 2
21
2 2lim
2x
x x x
x x→
+ − −+ −
2
6. 3
3 21
3 2lim
2 2 2x
x x
x x x→
− +− + −
0
7. 4 3 2
3 23
2 2 16 24lim
6x
x x x x
x x x→−
− − ++ −
10−
8. 4 3 2
3 20
2 2 16 24lim
6x
x x x x
x x x→
− − ++ −
4−
9. limm m
x a
x a
x a→
−−
m 1m a −⋅
10. limm m
n nx a
x a
x a→
−−
m nma
n−
11. 2
2 2lim
2x
x
x→
+ −−
1
4
12. 3
6 3lim
3x
x x
x>→
+ −−
0
13. 0
lim1 1x
x
x x→ + − − 1
14. 2 2
5
4 5 25lim
5x
x x x
x>→
− − − −−
−∞
15. 32
6 3 4 1lim
2x
x x
x→
− − +−
0
Calculus 15
16. 1
2 3 3lim
3 2x
x x
x→
− + ++ −
2
17. 3 23
22
1 1 2lim
2x
x x x
x x→
− + + −−
1
2−
18. 4
3
2 1lim
2 1x
x
x→
− −− −
1
2
19. 3 33 2 32
2 2lim
3 4 4x
x x
x x x x x><
→
− −− − + −
3
3±
20. Evaluate a so that
2 2 2
2 25
8 8lim
2 3 12x
x ax x a a
x ax a→
+ − + =+ −
a 2=
21. 2 21
2 3lim
3 2 1x x x x><
→
− − + − ∞∓
22. 2 22
8 6lim
6 2x
x
x x x x→
+ − + − − −
7
15
23. 23
2
12
3lim1
4 3
x
xx
xx
x x
→−
+++
++ +
4
9
24. 22 3 1
lim2x
x x
x→+∞
+ − +−
1−
25. 3 3 2
3
8 2 4lim
1x
x x x
x x→∞
+ − −+ +
1;3
26. 3 31 8
lim1x
x
x→∞
−+
2−
27. ( )2lim 3 4x
x x x→∞
− − +
±∞
28. ( )3 3 2lim 3 1x
x x x→∞
+ + −
1
29. ( )2lim 1 4 1x
x x x→∞
+ − + +
−∞
Calculus 16
30. ( )lim 1x
x x x→∞
+ −
1
2
31. ( )2 2lim 2 4 9 1 4x
x x x x→−∞
+ + + − −
+∞
32. 3 2
22
2 8 4lim 4 5 1
4x
x x xx x
x→∞
− − + − + − −
9;
4−∞ −
33. ( )2 2lim 5 8 5 4x
x x x x x→∞
+ + − + −
6±
Calculus 17
5 Differentiation
5.1 The derivative in a point
Consider:
1 1
0 0
( ) ( )lim limx x
f f x x f x
x x∆ → ∆ →
∆ + ∆ −=∆ ∆
If this limit exists, in other words, if 0
limx
fIR
x∆ →
∆ ∈∆
, then we call the limit the derivative of the
function f in the point x1. We say that the function f is differentiable or has a derivative in x1.
.
Notation: 10
'( ) limx
ff x
x∆ →
∆=∆
or 1( )Df x or 1( )df x
dx
5.2 Geometrical interpretation of the derivative
Consider the curve with equation y = f(x), and 2 neighboring points 1 1( , ( ))A x f x and
1 1( , ( ))B x x f x x+ ∆ + ∆ on that curve. Then the difference ratio 1 1( ) ( )f f x x f x
x x
∆ + ∆ −=∆ ∆
is the
slope of the line AB. When 0x∆ → then the line AB rotates around A and reaches the tangent
at the function in the point A.
Conclusion: 1 11
0
( ) ( )'( ) lim
x
f x x f xf x
x∆ →
+ ∆ −=∆
is the slope of the tangent at f in the point
1 1( , ( ))A x f x .
This means that the equation of the tangent line at 1 1( , ( ))x f x of the function y = f(x) is given
by:
1 1 1( ) '( )( )y f x f x x x− = −
A
B
1x 1x x+ ∆ x
y
1( )f x x+ ∆
1( )f x f∆
x∆
( )y f x=
Calculus 18
Example
2
0
2 2
0
0
3 4
(1 ) (1)'(1) lim
(1 ) 3(1 ) 4 (1 3 1 4)lim
lim( 1) 1
x
x
x
y x x
f x ff
x
x x
xx
∆ →
∆ →
∆ →
= − +
+ ∆ −=∆
+ ∆ − + ∆ + − − ⋅ +=∆
= ∆ − = −
5.3 Left-hand and right-hand derivative in a point
One can consider in0
limx
f
x∆ →
∆∆
separately the right-hand limit and the left-hand limit. If they
exist, we call:
1 11
0 0
( ) ( )lim lim ( )x x
f x x f xff x
x x> >
+∆ → ∆ →
+ ∆ −∆ ′= =∆ ∆
the right-hand derivative in1x ,
1 11
0 0
( ) ( )lim lim ( )x x
f x x f xff x
x x< <
−∆ → ∆ →
+ ∆ −∆ ′= =∆ ∆
the left-hand derivative in1x .
A function is differentiable in x1 if and only if the left-hand derivative and the right-hand
derivative in x1 are equal.
If both one-sided derivatives in x1 exist but they are different, the function is not differentiable.
This means that there exist two tangent lines in x1. A point like that is called a corner point.
Example
211
4y x= − is continuous for each value of x, buth the right-hand derivative in x = 2, is
different from the left-hand derivative in that point, namely
2
2
lim '( ) 1
lim '( ) 1
x
x
f x
f x>
<
→
→
=
= −
Conclusion:
If f is continuous in x = x1 ⇒ f is differentiable in x = x1.
But one can prove that, if f is differentiable in x = x1 ⇒ f is continuous in x = x1.
Calculus 19
5.4 A vertical tangent
Example
The function 3y x= is continuous for each value of x.
We calculate the derivative of f in x = 0:
3
3 20 0 0
(0 ) (0) 1lim lim limx x x
f x f x
x x x∆ → ∆ → ∆ →
+ ∆ − ∆= = = +∞∆ ∆ ∆
So the function is not differentiable in x=0. We call this an undefined derivative. Geometrically
this means that the function has a vertical tangent in 0.
5.5 Differentiation rules
To calculate derivatives it is advisable to use the following differentiation rules rather than
using the limit-definition.
Let f and g be two functions that are differentiable in x.
1. the derivative of a constant function c
' 0c =
2. the derivative of the argument
' 1x =
3. the derivative of a som of functions ( ) '( ) '( ) '( )f g x f x g x+ = +
4. the derivative of a product of functions ( ) '( ) '( ) ( ) ( ) '( )f g x f x g x f x g x⋅ = ⋅ + ⋅
5. the derivative of a quotient of functions
'
2
'( ) ( ) ( ) '( )( ) if ( ) 0
( )
f f x g x f x g xx g x
g g x
⋅ − ⋅= ≠
6. the derivative of a power of a function
( )' 1( ) ( ) '( )n nf x n f x f x−= ⋅ ⋅
Calculus 20
7. the derivative of a composite function: the chain rule ( ) '( ) '( ( )) '( )g f x g f x f x= ⋅�
9. the derivative of trigonometric functions
2
2
(sin ) ' cos
(cos ) ' sin
1(tan ) '
cos1
(cot ) 'sin
x x
x x
xx
xx
== −
=
= −
Calculus 21
5.6 Exercises
Calculate the derivative of the following functions:
1. 3 2 1y x x x= − − + 2' 3 2 1y x x= − −
2. 3 21 34
8 4y x x= + − 23 3
'8 2
y x x= +
3. ( )22 4 3y x x= − + ( )( )2' 2 4 3 2 4y x x x= − + −
4. ( ) ( )23 6 3y x x x= + + ( )( )2 2' 3 6 2 13 18y x x x x= + + +
5. 2
2,51
5 32
yx x
=− + −
22
2,5 7,5'
15 3
2
xy
x x
−= − + −
6. 2
2
2 3
2 2
xy
x x
−=− +
( )
2
22
4 14 6'
2 2
x xy
x x
− + −=− +
7. 2 5 7
2
x xy
x
− +=−
( )2
2
4 3'
2
x xy
x
− +=−
8. 2
( 2)1
x xy
x
+=−
( )
2
22
2( 1)'
1
x xy
x
− + +=−
9. 31 23
y x xx
= + + 22
2' 1y x
x= + −
10. 3
2 1
xy
x=
−
( )( )2 2
22
3'
1
x xy
x
−=
−
11. 2 2y a x= − 2 2
'x
ya x
−=−
12. 218y x x= + −
2
2
18'
18
x xy
x
− −=−
13. 3 23y x x= +
2
3 2
3 6'
2 3
x xy
x x
+=+
14. 32 5 5y x x= + −
2
3
6 5'
2 2 5 5
xy
x x
+=+ −
Calculus 22
15. 24 1 2y x x x= − − − − 2
2 1' 4
2 2
xy
x x
−= −− −
16. ( ) 22 1 4y x x= − − 2
2
(2 1)' 2 4
4
x xy x
x
−= − −−
17. ( )2 21 4 3y x x x= + − + ( )( )2
2
1 3 9 4'
4 3
x x xy
x x
+ − +=
− +
18. 3 25 4 1y x x= − + ( )
( )223
2 5 2'
3 5 4 1
xy
x x
−=
− +
19. ( ) ( )32 245 6 4 2y x x x= − + +
3 2
24
35 30 52 24'
2 2
x x xy
x
− + −=+
20. 5x
yx
+= 2
10'
2 5
xy
x x
+= −+
21. 2
2
2 3 1
1
x xy
x x
− +=− + ( )
3 2
2 2
4 6 9 5'
2 1 1
x x xy
x x x x
− + −=− + − +
22. ( )5
3
1 2
xy
x=
+
( )( )7
3 1 3'
1 2
xy
x
−=
+
23. ( )( )( ) ( )
1 2
3 4
x xy
x x
− −=
− −
( )( )( ) ( )
2
3 3
2 10 11'
1 2 3 4
x xy
x x x x
− + −=− − − −
24. ( )2 2
2
2 3 9
4
x xy
x
− −=
4 2
3 2
3 2 36'
4 9
x xy
x x
− − +=−
25. ( )
( )
2 3 3
3
2 1 2
2
x xy
x
− −=
+
( )( )( ) ( )
3 2
24 33
2 1 14 2 4 22'
2 2
x x x xy
x x
− − + −=
+ −
26. 24siny x= ' 4sin 2y x=
27. cos 2 5cos 2y x x= − − ' 2sin 2 5siny x x= − +
28. 2cos2
xy = ' sin cos
2 2
x xy = − ⋅
29. sin cos
sin cos
x xy
x x
+=−
( )2
2'
sin cosy
x x
−=−
Calculus 23
30. siny x x= − ' 1 cosy x= −
31. sin 2 2siny x x= − ' 2cos2 2cosy x x= −
32. cosy x=
sin'
2 cos
xy
x
−=
33. 3 2cos 4y x= 3
8sin 4'
3 cos4
xy
x
−=
34. 4cos3 3sin 4y x x= + ' 12sin3 12cos4y x x= − +
35. cos sec 2y x x= + ' sin 2tan 2 sec2y x x x= − +
36. cos 2
sin
xy
x=
2
2sin sin 2 cos cos2'
sin
x x x xy
x
− ⋅ − ⋅=
37. 2 2
sin cos
3cos 2sin
x xy
x x
⋅=−
( )
4 4
2 2 2 2
3cos 2sin'
3cos 2sin 3cos 2sin
x xy
x x x x
+=− −
Calculus 24
6 Indefinite integrals
6.1 Antiderivative functions
Definition
Let us consider a function f: IR → IR : x → f(x).
An antidervative function F of the function f is any function with the property F’(x) = f(x).
Property 1
If F is an antiderivative of the function f, then F+k with k an arbitrary constant є IR, also is.
Property 2
If F1 and F2 are both antiderivatives of f, then they differ only in the constant.
Conclusion
If F is an antiderivative of f, then all antiderivatives of f can be found by adding to F an
arbitrary real constant.
6.2 The indefinite integral
The set of antidervatives of a function f is called the indefinite integral of f.
Notation:
{ }( ) ( ) | '( ) ( ) and k IRf x dx F x k F x f x= + = ∈∫
or
( ) ( )f x dx F x k= +∫
6.3 Properties
1. ( ) ( )k f x dx k f x dx=∫ ∫
2. ( )( ) ( ) ( ) ( )f x g x dx f x dx g x dx+ = +∫ ∫ ∫
Calculus 25
6.4 Basic integrals
1
, if n -11
nn x
x dx kn
+
= + ≠+∫
sin cosxdx x k= − +∫
cos sinxdx x k= +∫
2 tancos
dxx k
x= +∫
2 cotsin
dxx k
x= − +∫
6.5 Integration by substitution
Let f be a continuous function of x and x = g(t) is a differentiable function of t then:
( ) ( ( )) '( )f x dx f g t g t dt= ⋅∫ ∫
Calculus 26
6.6 Exercises
Calculate the indefinite integral of the following functions:
1. 3 x 33
4x x k+
2. 34 x 344
7x x k+
3. x x 22
5x x k+
4. 3 5 2x x 4 5 25
22x x k+
5. 3
1
x 3 23
2x k+
6. 9
1
x
8
1
8k
x− +
7. 34
3 2
x
x 1212
13x x k+
8. 2 7x+ 2 7x x k+ +
9. sin cosx x+ cos sinx x k− + +
10. 2
7
2cos x
7tan
2x k+
11. 1
1 cos2x−
1cot
2x k− +
12. 2
5
5 8 3x x
x
− +
2 3 4
5 8 3
2 3 4k
x x x− + − +
13. 3
2
1 cos
cos
x
x
− tan sinx x k− +
14. 1 cos2x+ 2 sinx k± +
15. ( )53x+
( )63
6
xk
++
16. cos2x 1
sin 22
x k+
Calculus 27
17. 2cos x 1 1
sin 22 4
x x k+ +
18. 2sin x 1 1
sin 22 4
x x k− +
19. ( )45x−
( )55
5
xk
−+
20. 6x+ ( )26 6
3x x k+ + +
21. 4
5x− 8 5x k− +
22. sin3x 1
cos33
x k− +
23. ( )52 7x −
( )62 7
12
xk
−+
24. 3 4x + ( )23 4 3 4
9x x k+ + +
25. 2
1
cos 4x
1tan 4
4x k+
26. 3
1
3 1x − ( )231
3 12
x k− +
27. ( )2 2x x− − ( )222 2
5x x k− − +
28. ( )3 2 6x x+ + ( )223 2 6
5x x k+ + +
29. 3sin x 3cos
cos3
xx k− + +
30. 3cos x 3sin
sin3
xx k− +
31. sin 2 cosx x⋅ 32cos
3x k− +
32. 4cos x 3 1 1
sin 2 sin 48 4 32
x x x k+ + +
Calculus 28
33. 4sin x 3 1 1
sin 2 sin 48 4 32
x x x k− + +
34. 1
1 cosx+ tan
2x
k+
35. 3x x+ ( )226 3
5x x x k+ − + +
36. 2 1x x− ( )( )221 15 12 8 1
105x x x x k− + + − +
37. ( ) 33 2x x+ + ( )( ) 332 4 15 2
28x x x k+ + + +
38. 2
3 1
x
x
+−
( )23 20 3 1
27x x k+ − +
Calculus 29
6.7 Integration by parts
Let u = f(x) and v = g(x) be differentiable functions.
Since ( )d u v u dv v du⋅ = +
it follows that ( )u dv d u v v du= ⋅ −
and therefore
u dv uv v du= −∫ ∫
6.8 Exercises
Evaluate the indefinite integral of the following functions:
1. 2 cosx x 2 sin 2 cos 2sinx x x x x k+ − +
2. cosx x sin cosx x k+ +
3. 3 sinx x 3 2cos 3 sin 6 cos 6sinx x x x x x x k− + + − +
4. 2cos x 1 1
sin cos2 2
x x x k+ +
5. 3cos x 21 2cos sin sin
3 3x x x k+ +
6. sin cosx x x 1 1
cos2 sin 24 8
x x x k− + +
7. 2cosx x 21 1 1sin 2 cos2
4 4 8x x x x k+ + +
Calculus 30
7 Definite integrals
7.1 The fundamental theorem of calculus
Let the function f be continuous on [a,b] and let F be an antiderivative of the function f on
[a,b] then
[ ]( ) ( ) ( ) ( ) ( ) |b
b baa
a
f x dx F b F a F x F x= − = =∫
Remark: The definite integral is a real number, the indefinite integral is a set of antiderivatives.
Example
2
0
sin 2I xdx
π
= ∫
1 1sin 2 sin 2 (2 ) cos2
2 2xdx xd x x k= = − +∫ ∫
Therefore [ ] ( )20
1 1cos2 1 1 1
2 2I x
π= − = − − − =
7.2 Properties of the definite integral
Let f and g be functions that are continuous on a finite interval [a,b] .
1. ( ) ( )a b
b a
f x dx f x dx= −∫ ∫
2. ( ) ( )b b
a a
k f x dx k f x dx=∫ ∫ with k a real constant.
3. ( )( ) ( ) ( ) ( )b b b
a a a
f x g x dx f x dx g x dx+ = +∫ ∫ ∫
Calculus 31
7.3 The substitution method
Given the integral ( )b
a
f x dx∫ and f is continuous on [a,b] .
In this method we apply the substitution rule to the integrand and convert the limits as well.
Example
2
0
sin 2I xdx
π
= ∫
let 2x = t then 2dx =dt
0 0
2
x t
x tπ π
= ⇒ = = ⇒ =
Therefore ( )00
sin 1 1cos cos cos0 1
2 2 2
t dtI t
ππ
π = = − = − − = ∫
7.4 Partial integration
If u = u(x) and v = v(x) are differentiable functions of x and let udv uv vdu= −∫ ∫ then
[ ]bb b
b
aa aa
u dv uv vdu u v v du = − = − ∫ ∫ ∫
Example
2
1
1I x x dx= −∫
Set ( )32
1 1 13
u x du dx
dv x dx v x dx x
= ⇒ = = − ⇒ = − = − ∫
⇒ ( ) ( ) ( ) ( ) ( )22 2
3 3 5
1 1 1
2 2 2 2 2 4 4 161 1 2 0 1 1 0
3 3 3 3 5 3 15 15I x x x dx x
= − − − = − − ⋅ − = − − = ∫
Calculus 32
7.5 Exercices
Evaluate the following definite integrals:
1. 3
1
xdx∫ 4
2. ( )2
2
2
3 2 4x x dx−
− +∫ 32
3. 0
sinxdxπ
∫ 2
4. 2
21
1dx
x∫
12
5. 2
2cos xdxπ
π∫
2π
6. 1
2
0
1 x dx−∫ 4π
7. 1
5
5x x dx−
−
+∫ 20815
−
8. 2
22 5
xdx
x− +∫ 0
9. 2 3
0
cos sinx xdxπ
∫ 4
15
10. 2
3
0
cos sinx xdx
π
∫ 14