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