Lycée Denis-de-Rougemont 2006 – 2007 TABLE OF CONTENTSLycée Denis-de-Rougemont 2006 – 2007 - 2 - YAy & PRi § 0. REMINDER 0.1 Continuity The function f is continuous at x = a

3MG Level 1 CALCULUS II Theory Lycée Denis-de-Rougemont 2006 – 2007

- 1 - YAy & PRi

TABLE OF CONTENTS

§ 0. REMINDER .............................................................. 2

0.1 Continuity..........................................................................2

0.2 Derivative of a function .........................................................2

§ 1. EXPONENTIAL AND LOGARITHMIC FUNCTIONS........... 5

1.0 Euler’s number ....................................................................6

1.1 Special bases .......................................................................6

1.2 Asymptotic behaviour............................................................7

1.2.1 Logarithmic functions ..................................................................... 7

1.2.2 Exponential functions...................................................................... 8

1.2.3 Match (conflict) exponentials – logarithms – polynomials ........................10

1.3 Differentiating exponential and logarithmic functions ................. 14

1.3.1 Differentiating exponential functions .................................................14

1.3.2 Differentiating exponential to base b (b ≠ e) .........................................15

1.3.3 Differentiating logarithmic functions .................................................16

§ 2. INTEGRATION.........................................................16

2.0 Riemann’s sums or rectangle method....................................... 16

2.1 Fundamental Theorem of Calculus ......................................... 19

2.2 Integration of special functions .............................................. 21

2.3 Integration by parts............................................................. 25

2.4 Area calculus..................................................................... 26

2.5 Mean of a function ............................................................. 29


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§ 0. REMINDER

0.1 Continuity

The function f is continuous at x = a if lim ( ) ( )x a

f x f a→

= . It means that the limit when x

tends to a from the left and from the right must be equal to ( )f a .

The function f is continuous on an interval if it is continuous at each value of the

interval.

The points of discontinuity of a function are represented by vertical asymptotes, jumps

or holes.

0.2 Derivative of a function

Given the function ( )y f x= . The derivative of the function f is a function named '( )f x

and is defined by :

0

( ) ( )'( ) lim

x

f x x f xf x

x∆ →

+ ∆ −=

∆

Provided that limit exists. This expression is the limit of the differential quotient.

Geometrical interpretation

The derivative of f evaluated at x a= (that is to say '( )f a ) represents the gradient of

the tangent to the curve f at the point with abscissa x a= .

Definitions

1. A function is differentiable at the point ( );p p

P x y if its derivative is defined at P,

that is to say if '( )pf x ∈� .

2. A function is differentiable on an interval if it is differentiable at each point of

that interval.

Result

A function which is differentiable on an interval is also continuous on that interval.


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Rules of derivation

We have some rules available to calculate the derivative of a function : multiplying by a

constant, sum, product, quotient and chain rule :

1. Derivative of the powers of x

1( ) then '( ) ,n nf x x f x nx n−= = ∈�

2. The sum rule

( ) ( ) ( ) then '( ) '( ) '( )f x u x v x f x u x v x= + = +

3. The multiplying by a constant and product rule

( ) ( ) then '( ) '( )f x k u x f x k u x= ⋅ = ⋅

( ) ( ) ( ) then '( ) '( ) ( ) ( ) '( )f x u x v x f x u x v x u x v x= ⋅ = ⋅ + ⋅

4. The quotient rule

2

( ) '( ) ( ) ( ) '( )( ) then '( )

( ) ( ( ))

u x u x v x u x v xf x f x

v x v x

⋅ − ⋅= =

5. The chain rule

( ) ( ( )) then '( ) '( ( )) '( )f x u v x f x u v x v x= = ⋅

Examples

1. The multiplying constant rule

3( ) 25f x x= then 2 2'( ) 25 3 75f x x x= ⋅ =

2. The sum rule

4( ) 3 cos( )f x x x= + then 3'( ) 12 sin( )f x x x= −

3. The multiplying rule

( ) sin( )f x x x= ⋅ then '( ) 1 sin( ) cos( ) sin( ) cos( )f x x x x x x x= ⋅ + ⋅ = +

4. The quotient rule

4

5( )

3

xf x

x

+= then

4 3 4 4 3

4 2 8 5

1 3 ( 5) 12 3 12 60 3 20'( )

(3 ) 9 3

x x x x x x xf x

x x x

⋅ − + ⋅ − − − −= = =


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5. The chain rule

a) ( ) cos(2 1)f x x= + then ( ) cos( )( ) 2 1

'( ) sin(2 1) 2 2sin(2 1)u x xv x x

f x x x== +

− + ⋅ = − +=

b) 5( ) ( 3 4)f x x= − + then 5

4 4

( ) ( )( ) 3 4

'( ) 5( 3 4) ( 3) 15( 3 4)u xv x x

f x x x==− +

− + ⋅ − = − − +=

Terminology

Critical point (CP) : point whose derivative is undefined or equal to zero.

Stationary point (STP): point such that '( ) 0f x = (horizontal tangent)

Turning point (TP) : Maximum, minimum or level point (type of a STP)

The type of a SP is obtained with the table of variations of the function :

Maximum : STP such that f’ is positive before and negative after

Minimum : STP such that f’ is negative before and positive after

Level point : STP and inflexion point (the sign of f’ doesn’t change)

Vertical tangent point (VTP) : point such that lim '( )x a

f x→

= ±∞

« Angular point » : point such that lim '( ) lim '( )x a x a

f x f x− +→ →

≠

Utilizations of the derivative

The optimization problems (research of the maximal or minimal value) are solved by

considering the derivative of the function that has to be optimized.

The extreme value of the function ( )y f x= is reached either at an endpoint of the

interval of the possible values, or at a critical point but generally the solution is a

stationary point.

The derivative is also useful to sketch the graph of a given function. Actually, thanks to

it, we can determine where the function is increasing or decreasing and we can locate

its stationary points.

Remark

The main calculus results are in the « Formulaires et Tables » on pages 65-77.


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§ 1. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

A function of the form ( ) , where 0, 1 andxf x b b b x= > ≠ ∈� is called an exponential

function to base b, because the variable x appears in the exponent. We have seen that

it has an inverse function called logarithm to base b :

log ( )x

by b x y= ⇔ =

As these two functions are inverse functions they are reflections of each other in the line

y x= . The domain of ( ) xf x b= is � and its range is *

+� (its graph is strictly positive).

Consequently, the domain of logarithmic functions is *

+� (logarithmic functions don’t exist

for negative values of x) and the range is � .

Illustration

Remark

The number log ( )b a is the answer to the question : To which power do I have to raise

the number b to obtain a ?

Properties of logarithms

log (1) 0b = for any 0, 1b b> ≠

log ( )n

b b n= for any n

log( ) log( )nx n x= The power rule

( ) 1log log( )n x x

n= The nth root rule

log( ) log( ) log( )pq p q= + The multiplication rule

log log( ) log( )p

p qq

= −

The division rule

expD = � *

expR += �

logR = � *

logD += �

xy b=

log ( )bx y=

x1

x3

x2

…

y1

y3

y2

… -2 2 4

-3

-2

-1

1

2

3

4

5

xy b=

log ( )by x=

y x=


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1.0 Euler’s number

Leonhard Euler (April 15, 1707 – September 18, 1783) was a Swiss mathematician and

physicist. He is considered to be one of the greatest mathematicians who ever lived.

Leonhard Euler was the first to use the term « function » to describe an expression

involving various arguments ; i.e., y = f(x). He is credited with being one of the first to

apply calculus to physics.

Born and educated in Basel, he was a mathematical child prodigy. He worked as a

professor of mathematics in St. Petersburg, later in Berlin, and then returned to St.

Petersburg. He is the most prolific mathematician of all time, his collected work filling 75

volumes. He dominated 18th century mathematics and deduced many consequences of

the newly invented calculus. He was almost completely blind for the last seventeen years

of his life, during which time he produced almost half of his total output.

The Euler’s number, denoted by the letter e, is defined by the following limit :

2,7182818284591

045235360287lim 4713...1

n

ne

n→∞

= + =

As the number π , e is an irrational number. If we calculate some approximations of e we

obtain :

10

100

1000

110 1 2,5937

10

1100 1 2,7048

100

11000 1 2,7169

1000

n e

n e

n e

= ≈ + ≈

= ≅ + ≅

= ≅ + ≅

1.1 Special bases

Although the base of the logarithmic functions can be any real positive number except 1,

in practice only two bases are in common use. The first one is the base e (the Euler’s

number). This special logarithm is labelled « ln » or « LN » on your calculator and named

natural logarithm. The other base is 10, which is important because our system of

writing numbers is based on powers of 10. We note this logarithm either by « Log »,

« log », or, on your calculator, « LOG ».


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It is useful to be able to express a logarithm to a given base in base e or 10. For that we

consider the following logarithm to base b :

log ( )b x y=

Then, as log ( )b x is the inverse function of the exponential xy b= we have the relation :

log ( ) y

b x y x b= ⇔ =

We use the logarithm to base c, with or 10c e c= = , on both sides of the equation yx b= :

log ( ) log ( )

log ( ) log ( ) by the power rule

log ( )log ( )

log ( )

y

c c

c c

cb

c

x b

x y b

xy x

b

⇔ =

⇔ =

⇔ = =

If we rewrite this formula with ln and log, we obtain :

log( ) ln( )log ( )

log( ) ln( )b

x xx

b b= =

Example

Compute 7log (5) . Thanks to this formula : 7

log(5) ln(5)log (5) 0,83

log(7) ln(7)= = ≅

1.2 Asymptotic behaviour

Reminder

The asymptotic behaviour of a function is the study of the behaviour of this function

close to its excluded values (vertical asymptotes or holes are possible) and for very

large ( x → +∞ ) or small values ( x → −∞ ) of x (may be a horizontal or a slant asymptote).

1.2.1 Logarithmic functions

Remark

As all logarithms are multiples of each other, the asymptotic behaviour is the same

for all logarithms thus we consider here the natural logarithm ln( )y x= . Actually, we

can write that :

�

ln( ) 1log ( ) ln( ) ln( ), where is a constant

ln( ) ln( )b

c

xx x c x c

b b= = ⋅ = ⋅


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The domain D of the natural logarithm is ] [ *0;D += ∞ = � . First of all, we study the

behaviour of ln( )y x= close to 0. For that we have to calculate the following limit :

0lim ln( )x

x+→

. If we try with numerical values we obtain :

x 0,1 0,01 0,001 … 10-1000 … 0

f(x) = ln(x) -2,30 -4,61 -6,91 … -2302,59 …

We deduce that 0

lim ln( )x

x+→

= −∞ and the vertical line 0x = is a vertical asymptote of ln( )y x= .

We still have to study the behaviour of f(x) = ln(x) for large values of x. One more time we

use numerical values :

x 10 100 1000 … 101000 …

f(x) = ln(x) 2,30 4,61 6,91 … 2302,59 …

We observe that lim ln( )x

x→+∞

= +∞ and there isn’t any slant or horizontal asymptote.

Graph of the natural logaritm

1.2.2 Exponential functions

Remark

As for logarithmic functions, the asymptotic behaviour is the same for all exponentials

and we consider here the exponential to base e shortly named exponential. Actually :

)ln( ln( ) ln( )

inverse property of rules offunctions logarithms indices

( )xx a x a x aa e e e= = =

ln( )y x=


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The domain D of xy e= is ] [;D = − ∞ + ∞ = � . Therefore there is no excluded value and

consequently no hole or vertical asymptote. We just have to study the behaviour of the

exponential for very large/small values.

For that we use numerical values :

x 2 4 10 …

f(x) = ex 7,38 54,6 22026,47 …

We observe that the exponential tends very fast to infinity so lim x

xe

→+∞= +∞ and there isn’t

any slant or horizontal asymptote.

Symmetrically we have :

x -2 -4 -10 …

f(x) = ex 0,14 0,01 54,5 10−⋅ …

The exponential tends very fast to zero so lim 0x

xe

→−∞= and the line 0y = is a horizontal

asymptote.

Graph of the exponential

xy e=


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To remember

1. Logarithms

If 0, then ( ) ln( ) vertical asymptote 0

If , then ( ) ln( ) no slant or horizontal asymptote

x f x x x

x f x x

>→ = → −∞ =

→ +∞ = → +∞

2. Exponentials

If , then ( ) 0 horizontal asymptote 0

If , then ( ) no slant or horizontal asymptote

x

x

x f x e y

x f x e

+→ −∞ = → =

→ +∞ = → +∞

1.2.3 Match (conflict) exponentials – logarithms – polynomials

The next question is : What is the asymptotic behaviour of the product of an exponential

and a logarithm or a logarithm and a polynomial function? Let’s consider each case

separately.

1. MATCH (CONFLICT) EXPONENTIAL – LOGARITHM

For example, we want to study the asymptotic behaviour of a function like

ln( )( ) ln( ) x

x

xf x x e

e

−= = ⋅ . The domain of this function is ] [0;D = + ∞ . First we

have to study the asymptotic behaviour close to zero. For that we calculate the

limit :

0

limit 0laws 0

lim ln( )ln( )lim

1limx

x xx

x

xx

e e

+

+

+

→

→

→

−∞= = = −∞

We deduce a vertical asymptote 0x = .

We still have to analyse the behaviour of the function when x → +∞ . In this

case, the limit above has to be calculated :

limit laws

lim ln( )ln( ) " "lim undetermined

limx

x xx

x

xx

e e→+∞

→+∞

→+∞

+ ∞= = =

+∞

We have to interpret this limit in order to give it a value. As we use to do for the

calculation of certain undetermined limits, we try with numerical values :

x 10 50 100 …

( ) ln( ) xf x x e−= 41,05 10−⋅ 227,54 10−⋅ 431,71 10−⋅ …


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Quickly the function f(x) tends to zero. Consequently ln( )

lim 0xx

x

e→+∞= .

The exponential decreases faster than the logarithm increases and we say that

the exponential wins against a logarithm. We observe this graphically too :

Illustration Watch the scale

And the function f(x) is :

ln( )( )

x

xf x

e=

xy e

−=

ln( )y x=


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2. MATCH (CONFLICT) EXPONENTIAL – POLYNOMIAL

If we consider a new “challenger” as polynomial functions we will have the same

situation. Let’s consider the function 10

( )xe

f xx

= . Its domain is *D = � . The study of

its asymptotic behaviour is :

Vertical asymptote (VA) or hole : 0

10 10limit0laws 0

lim " 1 "lim

lim 0

xx

x

x

x

ee

x x

−

−

−

→+→

→

= = = +∞ so VA 0x = .

Slant or horizontal asymptote (SA or HA) :

10 10limitlaws

lim 0lim 0

lim

xx

x

x

x

ee

x x→−∞

→−∞

→−∞

= = =+∞

so HA 0y =

10 10limitlaws

lim " "lim undetermined

lim

xx

x

x

x

ee

x x→+∞

→+∞

→+∞

+ ∞= = =

+∞

As before, we have to study more carefully this limit by using numerical values

or a graphic :

x 10 50 100 …

10( )

xef x

x= 62,20 10−⋅ 45,31 10⋅ 232,69 10⋅ …

Thanks to this table we can say that 10

limx

x

e

x→+∞= +∞ and there is a HA 0y = . The

exponential increases faster than the polynomial function. We say that the

exponential wins against a polynomial function.

10( )

x

ef x

x=


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3. MATCH LOGARITHM – POLYNOMIAL

The last question is what can we say about a match between the two losers?

Let’s study the asymptotic behaviour of the function ln( )

( )x

f xx

= . Its domain is

*D += � . The candidate for a vertical asymptote or a hole is 0x = . We calculate :

0

limit0laws 0

lim ln( )ln( ) " "lim

lim 0x

x

x

xx

x x

+

+

+

→+→

→

− ∞= = = −∞

The function has a VA : 0x = .

Then, when x → +∞ we have to calculate :

limitlaws

lim ln( )ln( ) " "lim undetermined

limx

x

x

xx

x x→+∞

→+∞

→+∞

+ ∞= = =

+∞

Numerical values give :

x 10 100 1’000 …

ln( )( )

xf x

x= 0,23 0,046 36,91 10−⋅ …

So ln( )

lim 0x

x

x→+∞= and the polynomial function wins against the logarithm.

ln( )( )

xf x

x=


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To remember : podium

1. Exponential functions win against logarithms.

2. Exponential functions win against polynomial functions.

3. Polynomial functions win against logarithms.

Remark

These results are only valid when x → ±∞ .

1.3 Differentiating exponential and logarithmic functions

1.3.1 Differentiating exponential functions

By definition of the derivative, we have, for the exponential function ( ) xf x e= :

0 0 rules of 0indices

0 limit 0laws

"0"undetermined

0

( ) ( )'( ) lim lim lim

( 1) ( 1)lim lim

x x x x x x

x x x

x x xx

x x

f x x f x e e e e ef x

x x x

e e ee

x x

∆ ∆

∆ ∆ ∆

∆ ∆

∆ ∆

∆∆ ∆ ∆

∆ ∆

+

→ → →

→ →

=

+ − − ⋅ −= = = =

⋅ − −= =

��

As a first result we can say that the derivative of the exponential is the exponential

itself multiplied by a constant 0

( 1)lim

x

x

ec

x

∆

∆ ∆→

−= . The problem is now to determine the

value of this constant.

Observation

If we replace by x x∆ , we obtain : 0

( 1)lim

x

x

ec

x→

−= . So we have to study the

asymptotic behaviour of the function 1

( )xe

f xx

−= when x tends to 0.

As in 1.2, we use numerical values :

x -0,1 -0,01 -0,001 0 0,001 0,01 0,1

1( )

xef x

x

−= 0,9516 0,9950 0,9995 1,0005 1,0050 1,0517


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We obtain 0

1lim 1

x

x

e

x→

−= and thanks to this result we can deduce that

0

( 1)lim 1

x

x

ec

x→

−= = .

Therefore, the derivative of the exponential is the exponential itself.

( ) xf x e= then '( ) xf x e=

Examples

Find the derivative of the following functions :

1. ( ) then '( ) ( ) ' chain rulex x xf x e f x x e e− − −= = − = −

2. 5 5 5( ) then '( ) (5 ) ' 5 chain rulex x xf x e f x x e e= = =

3. ( ) then '( ) ( ) ' ( ) ' (1 ) product rulex x x x x xf x xe f x x e x e e xe e x= = + = + = +

1.3.2 Differentiating exponential to base b (b ≠ e)

We have seen, in 1.2.2 that an exponential to base b can be written as :

ln( ) ln( )

inverse property offunctions logarithms

( )xx b x bf x b e e= = =

If we differentiate this equality, we obtain :

�ln( ) ln( )

chainrule

'( ) ( ) ' ( ) ' ln( ) ln( )x

x x b x b x

b

f x b e b e b b= = = =

Generalization

The derivative of an exponential to base b is the exponential to base b itself multiplied

by a constant equals to ln(b) :

( ) then '( ) ln( )x xf x b f x b b= =

Examples


1. ( ) 8 then '( ) ln(8) 8x xf x f x= = ⋅

2. 5 5 5

chain rule( ) 3 then '( ) (5 ) ' ln(3) 3 5ln(3) 3x x xf x f x x= = ⋅ = ⋅

3. prod. rule

( ) 7 then '( ) ( ) '7 (7 ) ' 7 ln(7) 7 7 (1 ln(7))x x x x x xf x x f x x x x x− − − − −= ⋅ = + = − ⋅ = −


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1.3.3 Differentiating logarithmic functions

As the natural logarithm is the inverse function of the exponential to base e, we have :

ln( )xx e=

If we differentiate each side of this equation we obtain :

�ln( ) ln( )

chain rule

1( ) ' ( ) ' 1 (ln( )) ' 1 (ln( )) ' (ln( )) 'x x

x

x e x e x x xx

= ⇒ = ⇒ = ⇒ =

So we have the result :

1( ) ln( ) then '( )f x x f x

x= =

Examples


1. chain rule

1 1( ) ln(2 ) then '( ) (2 ) ' (ln(2 )) ' 2

2f x x f x x x

x x= = ⋅ = ⋅ =

2. chain rule

1 1( ) ln( ) then '( ) ( ) ' (ln( )) 'f x ax f x ax ax a

ax x= = ⋅ = ⋅ =

3. prod. rule

1( ) ln( ) then '( ) ( ) ' ln( ) (ln( )) ' ln( ) ln( ) 1 f x x x f x x x x x x x x x x

x= ⋅ = ⋅ + ⋅ = ⋅ + ⋅ = +

4.

§ 2. INTEGRATION

2.0 Riemann’s sums or rectangles method

An important application of integration is to calculate areas and volumes. Many of the

formulas you have learnt so far, such as for the volume of a sphere or a cone, can be

proved by using integration. Integration is closely related to the differentiation process.

First of all we have to introduce the summation notation. Alternatively, the sum can be

represented by the summation symbol, which is the capital sigma Σ :

1 1...n

i m m n ni m

x x x x x+ −=

= + + + +∑


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The subscript gives the symbol for the index of summation i ; m is the lower bound of

summation, and n is the upper bound of summation. So, for example if we consider

1 2 101, 2, ... , 10, ...x x x= = = we have :

7

3 4 5 6 73

3 4 5 6 7ii

x x x x x x=

= + + + + = + + + +∑

Our next purpose is to calculate the area under a given

graph. More precisely, we want to determine the area

delimited by the x-axis, the curve ( )y f x= and the

vertical lines x a= and x b= .

That area A can be approximated by summing areas of

rectangles. We obtain an underestimation and an

overestimation. The more rectangles we use the better

the approximation will be. If we consider a continuous, positive function f on an

interval [ ];a b , we can calculate these two approximations of the area A described

above. For that, let’s consider the function f given by its graph. Then we divide up the

interval [ ];a b into n equal parts with equal length b a

nx∆ −= .

We name 1M the maximum and 1m the minimum of f on the first subdivision, 2M the

maximum and 2m the minimum on the second one, etc.

Illustration

A

b = xn a = x0

x x1 x2 xi-1 xn-1

Mi

xi

mi

1x∆

2x∆

ix∆

f


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We have for the overestimation and the underestimation of the area A :

Overestimation : 1 1 2 2 3 3

1

...n

n n i ii

A x M x M x M x M x M∆ ∆ ∆ ∆ ∆+

=

= ⋅ + ⋅ + ⋅ + + ⋅ = ⋅∑

Underestimation : 1 1 2 2 3 3

1

...n

n n i ii

A x m x m x m x m x m∆ ∆ ∆ ∆ ∆−

=

= ⋅ + ⋅ + ⋅ + + ⋅ = ⋅∑

Consequently we have the following inequalities :

1 1

n n

i i i ii i

x m A x M∆ ∆= =

⋅ ≤ ≤ ⋅∑ ∑

If we take x∆ smaller and smaller ( 0x∆ → ) we have :

0 01 1

lim limn n

i i i ix x

i i

x m x M A∆ ∆

∆ ∆→ →

= =

⋅ = ⋅ =∑ ∑

In the limit, the summation symbol Σ becomes ,∫ x∆ becomes dx and ( )i i iM m f x= = .

Finally the area is given by the formula :

( )b

a

A f x dx= ∫

( )b

a

f x dx∫ is called definite integral on the interval [ ];a b and is read « the integral of f

from a to b ». The sign∫ represents the integration, a and b are the endpoints of the

interval, f(x) is the function we are integrating and is called integrand. dx is a notation for

the variable of integration. Historically, dx represented an infinitesimal quantity, and the

long s stood for "sum".

Remark

Intuitively, the integral of a continuous, positive function f of one real variable x

between a left endpoint a and a right endpoint b represents the area bounded by the

lines x = a and x = b, the x-axis, and the curve defined by the graph of f.

Example

We consider the function ( ) ( 1)( 5)f x x x= − − − and we

want to calculate the enclosed area between its graph

and the x-axis. The exact value is A = 10,67.

Overestimations and the underestimations, using

10 and 25 rectangles, lead to the following results :

A = 10,67


- 19 - YAy & PRi

2.1 Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is the statement that the two central operations

of calculus, differentiation and integration, are inverses of each other. This means that if

a continuous function is first integrated and then differentiated, the original function is

retrieved. An important consequence of this allows one to compute integrals by using an

antiderivative of the function to be integrated.

Definition

An antiderivative of a function f(x) is a function, noted F(x), which satisfies the

condition '( ) ( )F x f x= .

Notation

We also use the following notation to name an antiderivative : ( )f x dx∫ . It is called

indefinite integral.

A = 8,96 A = 12,16

A = 11,29 A = 10,01

1 5

1

1

1 5

5

5


- 20 - YAy & PRi

Examples

1. ( ) 1 then ( )f x F x x c= = + 2. 2( ) 2 then ( )f x x F x x c= = +

3. 2 3 21( ) 4 then ( ) 2

3f x x x F x x x c= − = − + 4. 11

( ) then ( )1

n nf x x F x x cn

+= = ++

5. ( ) cos( ) then ( ) sin( )f x x F x x c= = + 6. 3 31( ) then ( )

3

x xf x e F x e c− −= = − +

Fundamental Theorem of Calculus

Given f a continuous function on the interval [ ];a b and F an antiderivative of f. Then

( ) ( ) ( )b

a

f x dx F b F a= −∫

Examples

Calculate the following definite integrals :

1. 3

4

00

3 4 4 4

2( ) 42 ( )

4 0 ( 2) 16f x x

F x x

x dx x−=

− =

= = − − = − ∫

2. 3

4

33

3 4 4 4

( ) ( 2)22 1

( ) ( 2)4

1 1 1 1 1( 2) ( 2) (3 2) (2 2) 0

4 4 4 4 4f x x

F x x

x dx x= −

= −

− = − = − − − = − =

∫

3. [ ]2

2

0( ) sin( )0 ( ) cos( )

sin( ) cos( ) cos(2 ) ( cos(0)) 0f x xF x x

x dx xπ

ππ

==−

= − = − − − =∫

4.

/ 4/ 4

( ) sin( )00 ( ) cos( )

1 1 1 1 1 1sin(4 ) cos(4 ) cos(4 ) cos(0)

4 4 4 4 4 4 2f x xF x x

x dx x

ππ π==−

= − = − ⋅ − − = + =

∫

Remark

In 2.0 we have said that, intuitively, the integral of a function represents the area under

its graph. But, if we consider the example 1, we have a negative area ! Consequently

we will have to study this kind of situations more carefully (see 2.4).


- 21 - YAy & PRi

Properties of the definite integral

1. ( ) ( )b b

a a

k f x dx k f x dx⋅ =∫ ∫

2. [ ]( ) ( ) ( ) ( )b b b

a a a

f x g x dx f x dx g x dx+ = +∫ ∫ ∫

3. ( ) 0a

a

f x dx =∫

4. ( ) ( )b a

a b

f x dx f x dx= −∫ ∫

5. ( ) ( ) ( )b c c

a b a

f x dx f x dx f x dx+ =∫ ∫ ∫

2.2 Integration of special functions

Since 2.1 we know how to calculate antiderivatives of easy functions as polynomials,

exponentials, etc. We are now going to learn how to calculate antiderivatives for more

complex functions.

1. FUNCTIONS OF THE FORM f ax + b( )

As integration is the inverse process of differentiation, we are going to find

antiderivatives of such functions by differentiating candidates for the antiderivative.

Let’s try with the antiderivative ( )F ax b+ . We have to check that

'( ) ( )F ax b f ax b+ = + . For that we use the chain rule :

: ( )

( ) ' ( ) '

': ' '( ) ' '( ) '( )

F x ax b v F v

F v a F v v F v a F ax b

+ =

↓ ↓

= ⋅ = ⋅ = ⋅ +

� �

The derivative is '

'( ) '( ) ( )F f

F ax b a F ax b a f ax b=

+ = ⋅ + = ⋅ + . Thus we have to correct

by dividing by 1

a. Finally, the antiderivative of ( )f ax b+ is :

1( ) then an antiderivative is ( )f ax b F ax b c

a+ + +


- 22 - YAy & PRi

Examples

1. 1

( ) sin(3 1) then ( ) cos(3 1)3

f x x F x x c= + = − + +

2. 1

( ) cos( 1) then ( ) ( cos( 1)) cos( 1)1

f x x F x x c x c= − + = ⋅ − − + + = − + +−

3. 9 10 101 1 1( ) ( 5 4) then ( ) ( 5 4) ( 5 4)

5 10 50f x x F x x x c= − + = − ⋅ − + = − − + +

2. FUNCTIONS OF THE FORM ⋅ue u'

Once it is understood that u is a function of x, it is very easy to integrate this kind

of functions. Actually, as the derivative of the exponent appears in the function we

deduce, thanks to the chain rule and 1.3.2, that :

( ) ( )( ) '( ) then ( )u x u xf x e u x F x e c= ⋅ = +

Justification

To verify this affirmation, we differentiate the antiderivative ( )( ) u xF x e c= + :

( ) �( ) ( ) ' ( )

chain0rule

'( ) ' '( ) '( )u x u x u xF x e c u x e c u x e OK=

= + = ⋅ + = ⋅ ⇒

Examples

1. 3 3( ) 3 then ( )x xf x e F x e c= = +

2. 3 32 2 4 2 4( ) ( 3 2) then ( )x x x xf x x e F x e c− + − − + −= − + = +

3. cos( ) cos( )( ) sin( ) then ( )x xf x x e F x e c= − = +

3. FUNCTIONS OF THE FORM ⋅f u x u' x( ( )) ( )

This kind of functions is actually the general case of 2. If we observe carefully the

form ( ( )) '( )f u x u x⋅ we recognize the chain rule, consequently an antiderivative is :

( ) ( ( )) '( ) then ( ) ( ( ))g x f u x u x G x F u x= ⋅ =


- 23 - YAy & PRi

Examples

1. ( ) 2sin(2 1) then ( ) cos(2 1)f x x F x x c= + = − + +

2. ( ) ( )1( ) cos then ( ) sin

2f x x F x x c

x= = +

4. FUNCTIONS OF THE FORM u'(x)

ax + b u(x)

1and

Since 1.3.3, we know that ( ) 1ln( ) 'x

x= we also know that for 0x >

1ln( )dx x c

x= +∫ .

We need the condition 0x > because the natural logarithm is only defined for

0x > . But the function 1

( )f xx

= exists for any x except 0.

Consequently we should be able to integrate this function over *� . To eliminate

this problem, we have to “transform” negative numbers into positive ones. For that

we use an absolute value, thus :

1

( ) then ( ) ln , 0f x F x x c xx

= = + ≠

It is now possible to find an antiderivative for functions as '( )

( )( )

u xf x

u x= and

1( )f x

ax b=

+.

One more time we use the chain rule : as the derivative of the function ( )u x

appears at the numerator we deduce :

'( )

( ) then ( ) ln ( ) , 0( )

u xf x F x u x c x

u x= = + ≠

If the function is of the form 1

( )f xax b

=+

we have :

1 1

( ) then ( ) lnf x F x ax b cax b a

= = + ++

(except if /x b a= − )


- 24 - YAy & PRi

Justification

To verify these results, we differentiate the antiderivative ( ) ln ( )F x u x c= + and

1( ) lnF x ax b c

a= + + :

( ) �'

chain0rule

1 1 '( )'( ) ln ( ) ' '( ) '( )

( ) ( ) ( )

u xF x u x c u x c u x OK

u x u x u x=

= + = ⋅ + = ⋅ = ⇒

'1 1 1 1

'( ) lnF x ax b c aa a ax b ax b

OK

= + + = ⋅ ⋅ = + +⇒

Examples

1. 1

( ) then ( ) ln 22

f x F x x cx

= = + ++

2. 1

( ) then ( ) lnf x F x x a cx a

= = + ++

3. 1 1

( ) then ( ) ln 5 15 1 5

f x F x x cx

= = − +−

4. 6

( ) then ( ) ln 6 76 7

f x F x x cx

= = + ++

5. 2

2

10 1( ) then ( ) ln 5 2

5 2

xf x F x x x c

x x

−= = − + +

− +

5. FUNCTIONS OF THE FORM 2 x(ax + bx + c)e

If we derivate this kind of functions, we obtain :

( )2 2'( ) (2 ) ( ) (2 ) ( )x x xf x ax b e ax bx c e ax a b x b c e= + + + + = + + + +

By stating that 2d a b= + and f b c= + we obtain : ( )2'( ) xf x ax dx f e= + + .

Consequently the derivative of such a function is a function of the same form. To

find an antiderivative, we identify the coefficients. Let’s consider an example.


- 25 - YAy & PRi

Example

( )2( ) 3 2 1 xf x x x e= − + . Find ( )F x .

We know that ( )F x is a function of the form ( )2( ) xF x ax bx c e= + + . We also

know that '( ) ( )F x f x= (by definition of an antiderivative). Thus :

( ) ( ) ( ) ( )2 2 2'( ) 2 (2 ) ( ) 3 2 1x x x xF x ax b e ax bx c e ax a b x b c e x x e= + + + + = + + + + = − +

We have the following conditions for the coefficients a, b and c :

3

2 2 3, 8, 9

1

a

a b a b c

b c

=

+ = − ⇒ = = − =+ =

and ( )2( ) 3 8 9 xF x x x e= − +

2.3 Integration by parts

There’s still one type of functions that we don’t know how to integrate. For instance, we

can’t find an antiderivative of the function ( ) xf x xe= . More generally, functions of that

kind, that is to say ( ) '( ) ( )f x u x v x= ⋅ (product of functions) can’t be integrated. The

process of finding antiderivatives of such functions is called integration by parts.

Integration by parts

[ ]'( ) ( ) ( ) ( ) ( ) '( )b b

b

aa a

u x v x dx u x v x u x v x dx⋅ = ⋅ − ⋅∫ ∫

Proof

The derivative of ( ) ( )u x v x⋅ is ( )( ) ( ) ' '( ) ( ) ( ) '( )u x v x u x v x u x v x⋅ = ⋅ + ⋅ thus

( ) [ ]( ) ( ) ' ( ) ( ) '( ) ( ) ( ) '( ) '( ) ( ) ( ) '( )b b b b

b

aa a a a

u x v x dx u x v x u x v x u x v x dx u x v x dx u x v x dx⋅ = ⋅ = ⋅ + ⋅ = ⋅ + ⋅∫ ∫ ∫ ∫

So [ ]'( ) ( ) ( ) ( ) ( ) '( )b b

b

aa a

u x v x dx u x v x u x v x dx⋅ = ⋅ − ⋅∫ ∫


- 26 - YAy & PRi

Examples

1. Find an antiderivative of the function ( ) xf x xe= .

We have to choose the functions '( )u x and ( )v x . As '( )u x has to be integrated

and ( )v x differentiated, we state that '( ) xu x e= ( xe is very easy to integrate) and

( )v x x= (the derivative of x is very simple). We obtain ( ) xu x e= and '( ) 1v x = , so

( ) 1 ( 1)

x

x x x x x x x x

e

F x e x dx e x e dx e x e xe e c e x c

= ⋅ = ⋅ − ⋅ = ⋅ − = − + = + + ∫ ∫��

2. Find an antiderivative of the function ( ) ln( )f x x x= .

We choose '( )u x x= and ( ) ln( )v x x= because it isn’t easy to integrate ln( )x . Then :

2

2 2 2 2

1 1

2 4

2 2 2

1 1 1 1 1( ) ln( ) ln( ) ln( )

2 2 2 4

1 1 1 1ln( ) ln( )

2 4 2 2

xdx x

F x x x dx x x x dx x x xx

x x x c x x c

=

= = − ⋅ = − =

∫

= + + = + +

∫ ∫��

2.4 Area calculus

We have seen that, for a positive continuous function, the integral ( )b

a

f x dx∫ is the area

bounded by the graph of f and the two vertical lines x a= and x b= . But what can we

say if the function isn’t positive ?

Using the definition (Riemann’s sums) of ( )b

a

f x dx∫ we interpret it as the signed area

enclosed by the x-axis, the curve and the verticals x a= and x b= .

Important

The area above the x-axis is considered as positive and the area below the x-axis is

considered as negative.

Example

Calculate the area under the curve y x= from -3 till 2.

22

2 2 2

33

1 1 12 ( 3) 2 4,5 2,5

2 2 2x dx x

−−

= = ⋅ − ⋅ − = − = −

∫


- 27 - YAy & PRi

As an area is always positive, that negative answer can’t be the value of the area

enclosed by the line y x= , the x-axis, and the verticals 3x = − and 2x = .

The area is the sum of two below triangles’ area and is equal to 2 4,5 6,5+ = . We

calculate it like that :

0 22 0 2

2 2

property3 03 3 0

1 14,5 2 6,5

2 2x dx x dx x dx x x

−− −

= + = + = − + =

∫ ∫ ∫

AREA UNDER A CURVE

A first application of integration that is directly linked to the definition of ( )b

a

f x dx∫ is the

determination of the area enclosed by the curve, the x-axis and the verticals x a= and

x b= . To solve that, we use the following 3 steps process :

Steps

1. Find all the x-intercepts (zeros) of the function ( )f x in [ ];a b : 1 2, ,..., nx x x

The signed area is likely to change its sign at these points.

2. Find the areas between the successive x-intercepts by calculating the definite

integrals 1

( )

x

a

f x dx∫ , 2

1

( )

x

x

f x dx∫ , … , and ( )

n

b

x

f x dx∫ .

3. Add the absolute values of the signed areas found at the previous step.

The definite integral’s result is

the sum of the signed areas.

+

–

1 4,5A =

2 2A =

( )f x x=


- 28 - YAy & PRi

Example

Let’s determine the area A between the curve 3y x x= − and the x-axis over

[ ]1,2;1,3− .

1. Zeros of ( )3 2( ) 1 ( 1)( 1)f x x x x x x x x= − = − = − + � 1 2 3( 1;0) (0;0) (1;0)I I I−

2. 1

1

1,2

( ) 0,0484A f x dx−

−

= = −∫

0

2

1

( ) 0,25A f x dx−

= =∫

1

3

0

( ) 0,25A f x dx= = −∫

1,3

4

1

( ) 0,119025A f x dx= =∫

3. 1 2 3 4 0,667425A A A A A= + + + =

AREA BETWEEN TWO CURVES

The area bounded by two continuous curves ( )f x and ( )g x is defined by the

intersection points of these curves. Then, the absolute values of the definite integrals

must be added.

Steps

1. Find the abscissas 1 2, ,..., nx x x of all the intersection points of ( )f x and ( )g x .

These are the solutions of the equation ( ) ( )f x g x= , so the zeros of ( ) ( )f x g x− .

2. Find the signed areas 2

1

1 ( ) ( )

x

x

A f x g x dx= −∫ , 3

2

2 ( ) ( )

x

x

A f x g x dx= −∫ ,…

3. Add the absolute values of the signed areas found in step 2 : 1 2 ... nA A A A= + + +

Example

Find the area between the curves 3 2( ) 3 1f x x x= − + − and ( ) 1,25 1g x x= − .

1. Zeros

3 2

3 2

2

3 1 1,25 1

3 1,25 0

( 3 1,25) ( 0,5)( 2,5) 0

x x x

x x x

x x x x x x

− + − = −

− + − =

− − + = − − − =

�

1

2

3

0

0,5

2,5

x

x

x

=

= =

A1

A4

A3

A2


- 29 - YAy & PRi

2. 3 2( ) ( ) ( ) 3 1,25h x f x g x x x x= − = − + − � 4 3 21

4

5( ) ( ) ( )

8H x F x G x x x x= − = − + −

10,5

4 3 2

1

00

1

4

5( ) ( ) 0,047

8A f x g x dx x x x

= − = − + − ≅ − ∫

2,52,5

4 3 2

2

0,50,5

1

4

5( ) ( ) 2

8A f x g x dx x x x

= − = − + − = ∫

3. 1 2 2,05A A A= + ≅

Variation

« Find the area A between the curves ( )y f x= and ( )y g x= over the interval [ ];a b ».

The only difference in the calculations is that we may have to add 1

( ) ( )

x

a

f x g x dx−∫

and ( ) ( )

n

b

x

f x g x dx−∫ .

2.5 Mean of a function

The integral between a and b of a function represents a signed area. Let’s consider the

following functions :

What must the height of a rectangle be so that its area is equal to the signed area

between the curve and the x-axis ?

I1

I2

I3

A1

A2


- 30 - YAy & PRi

Mathematically, we look for h such that ( ) ( )b

a

b a h f x dx− ⋅ = ∫ . This value is given by

1( )

b

a

h f x dxb a

= ⋅− ∫ . It is noted f an is called the mean of the function f on the interval [ ];a b :

1( )

b

a

f f x dxb a

=− ∫

For a continuous function f, the mean is one the values of f. Actually, there exists a

number [ ];c a b∈ such that ( )f f c= . This result is known as the Mean Value Theorem.

Example

Find the mean of 2( )f x x= over ] [2;2− .

22

2 3

22

1 1 1 1 1 1 48 ( 8)

2 ( 2) 4 3 4 3 3 3f x dx x

−−

= = = ⋅ − ⋅ − = − −

∫

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