International Baccalaureate
MATHEMATICS
Analysis and Approaches (SL and HL)
Lecture Notes
Christos Nikolaidis
TOPIC 5
CALCULUS
5.1
THE LIMIT limf(x) – THE DERIVATIVE f (x) : A ROUGH IDEA! ….……..
1
5.2
DERIVATIVES OF KNOWN FUNCTIONS – RULES ……………………………….
9
5.3
TANGENT LINE – NORMAL LINE AT SOME POINT x0 ………………………. 16
5.4
THE CHAIN RULE ………………………………..………………………………………………… 21
5.5
MONOTONY - MAX, MIN ……………………………………………………………………... 29
5.6
CONCAVITY - POINTS OF INFLEXION …………………….………………………..
34
5.7
OPTIMIZATION ……………………………….……………………………….…………………….
39
5.8
THE INDEFINITE INTEGRAL
5.9
INTEGRATION BY SUBSTITUTION …………………………………...………….………. 51
5.10
THE DEFINITE INTEGRAL
5.11
KINEMATICS (DISPLACEMENT, VELOCITY, ACCELARATION) …………… 72
 f(x)dx …………………………………………………… 44
b
 f(x)dx - AREA BETWEEN CURVES …….. 59
a
Only for HL
5.12
CONTINUITY AND DIFFERENTIABILITY ………………………………………………… 81
5.13
L’HÔPITAL’s RULE …………………………………………………………………………………... 90
5.14
IMPLICIT DIFFERENTIATION – MORE KINEMATICS ………………………….. 95
5.15
RATE OF CHANGE PROBLEMS ……………………………………………………………. 100
5.16
FURTHER INTEGRATION BY SUBSTITUTION ……………………………………… 105
5.17
INTEGRATION BY PARTS .……………………………………………………………………. 112
5.18
FURTHER AREAS BETWEEN CURVES - VOLUMES ………………………….. 118
5.19
DIFFERENTIAL EQUATIONS ………………………………………………………………… 123
5.20
MACLAURIN SERIES – EXTENSION OF BINOMIAL THEOREM ….…….. 134
November 2021
TOPIC 5: CALCULUS
5.1
Christos Nikolaidis
THE LIMIT limf(x) – THE DERIVATIVE f ΄(x): A ROUGH IDEA
This paragraph may look very “technical”. Do not pay much attention on
your first reading. You may skip and proceed to paragraph 5.2; you will
realize that the derivative in practice is much easier than it appears here!
 THE LIMIT limf(x)
Consider the function f(x) = 2x+3
It is clear that when x=2, then f(2) = 7.
But what is the behavior of f(x) when x approaches 2?
approach x=2
approach x=2
from values less than 2
from values greater than 2
x
f(x)
x
f(x)
1.8
6.6
2.2
7.4
1.9
6.8
2.1
7.2
1.99
6.98
2.01
7.02
1.999
6.998
2.001
7.002
that is
that is
if x 2= then f(x)  7
if x 2+ then f(x)  7
Thus in general
if
x 2
[x tends to 2]
then f(x)  7
[f(x) tends to 7]
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TOPIC 5: CALCULUS
Christos Nikolaidis
In order to express this fact we write
lim f(x)  7
x 2
and say that the limit of f(x), as x tends to 2, is 7.
Remark
In fact for the right column we write lim- f(x)  7
x 2
while for the right column we write lim f(x)  7
x 2
and these are called side limits. If the side limits are equal then
lim f(x)  7
x 2
In this example
lim f(x)  7  f(2)
x 2
lim f(x)  9  f(3)
and so on!
x 3
The situation lim f(x)  f(a) occurs very often, however, this is not
x a
always the case (otherwise the limit would be nothing more than a
simple substitution!).
Let’ s see a case where the limit is not a simple substitution!
We will find informally (by using our GDC) the limit
sinx
.
x 0
x
lim
Notice that the function is not defined at x=0. The graph looks like
y
1
x
-5
5
2
TOPIC 5: CALCULUS
Christos Nikolaidis
Let’s approach x=0 by using our GDC:
approach x=0-
approach x=0+
(from values less than 0)
(from values greater than 0)
x
f(x)
x
f(x)
-0.1
0.998334
0.1
0.998334
-0.01
0.999983
0.01
0.999983
-0.001
0.999999
0.001
0.999999
that is
that is
if x 0= then f(x)  1
if x 0+ then f(x)  1
We say that the limit when x tends to 0 is 1 and we write
lim
x 0
sinx
1
x
Sometimes the limit can be +∞ or -∞.
Let us investigate informally (by using our GDC) the limit lim
x 0
f(x)=
1
x
3
1
.
x
TOPIC 5: CALCULUS
Christos Nikolaidis
approach x=0-
approach x=0+
(from values less than 0)
(from values greater than 0)
x
f(x)=
1
x
x
f(x)=
1
x
-0.1
-10
0.1
10
-0.001
-1000
0.001
1000
-0.000001
-1000000
0.000001
1000000
Here we only have side-limits:
lim
x 0 
1
=+  and
x
lim
x 0 _
1
=- 
x
Remember:
If lim f(x)   or lim f(x)   we say x=a is a vertical asymptote.
x a
x a
Thus, in our example, x=0 is a vertical asymptote.
We also define limits of the form lim f(x) or lim f(x) . Thus, we
x  
x  
study the behavior of the function when x approaches +  or -  .
Let us find informally (by using our GDC) the limits lim
x  
1
1
, lim .
x


x
x
x approaches - 
x approaches + 
x
f(x)
x
f(x)
-1000
-0.001
1000
0.001
-1000000
-0.000001
1000000
0.000001
Both limits are 0, namely
1
=0
x   x
lim
and
1
= 0 .
x   x
lim
Remember:
If lim f(x)  a or lim f(x)  a then y=a is a horizontal asymptote.
x  
x - 
Thus, in our example, y=0 is a horizontal asymptote.
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TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 1
y
5
4
3
2
x 3
f(x) 
x2
1
-2
-1
x
1
2
3
4
5
6
-1
-2
-3
We know that

x=2 is a vertical asymptote.
The formal explanation is that

lim f(x)   and lim f(x)  
x 2
x 2
y=1 is a horizontal asymptote.
The formal explanation is that
lim f(x)  1 and lim f(x)  1
x  
x - 
Look at an interesting limit that provides the irrational number
e=2.7182818…
EXAMPLE 2
x
1

Investigate informally (by using your GDC) the limit lim  1   .
x   
x
x approaches + 
x
f(x)
1000
2.7169239…
1000000
2.7182804…
1010
2.7182818…
The resulting limit is in fact the number e=2.7182818… That is,
x
1

lim  1   = e .
x   
x
5
TOPIC 5: CALCULUS
Christos Nikolaidis
 RATE OF CHANGE (OR GRADIENT) IN A STRAIGHT LINE
Consider the line f(x)=2x+3.
y
8
7
6
5
4
3
2
1
-4
-3
-2
x
-1
1
2
3
4
-1
-2
Notice that
when
x changes from 1 to 2
then
y changes from 5 to 7
Hence, the corresponding rate of change is
Δy
Δx
=
f(2)  f(1)
7 5
=
= 2
2 1
2 1
We understand that the rate of change between any two points on
the line is always the same.
For example,
when
x changes from 0 to 2
then
y changes from 3 to 7
Hence, the corresponding rate of change is still
Δy
Δx
=
f(2)  f(0)
7 3
=
= 2
20
20
This common value is the gradient of the line.
Next, we will see that the gradient is not only defined for straight
lines but also for other curves.
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TOPIC 5: CALCULUS
Christos Nikolaidis
 RATE OF CHANGE (OR GRADIENT) ΙΝ Α CURVE
In a curve which is not a straight line, the rate of change between
any two points is not always the same.
For example, in f(x)=x2
Δy
Rate of change from x=1 to x=2:
Δx
Δy
Rate of change from x=1 to x=3:
Δx
=
f(2)  f(1)
4 1
=
3
2 1
2 1
=
f(3)  f(1)
91
=
4
31
31
However, we can measure the “instantaneous” rate of change at
any point on the curve. This will be the gradient at this point. In
the following diagram we provide the gradient at some points:
6
gradient at x=0: m =0
gradient at x=1: m =2
gradient at x=2: m =4
4
gradient at x=3: m =6
gradient at x=-1: m =-2
2
-2
0
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TOPIC 5: CALCULUS
Christos Nikolaidis
For example, the gradient at x=1 is m=2.
Justification:
For x=1, y=1. We will find the gradient at point A(1,1).
We select a neighboring point B at x=1+h (where h is very small)
x-coordinate = 1+h
y-coordinate =(1+h)2
The rate of change from point A(1,1) to point B(1+h,(1+h)2) is
Δy
Δx

(1  h)2 - 1 1  2h  h 2 - 1 2h  h 2


 2h
(1  h)- 1
h
h
If we let h become very small, that is h0, the result will be 2:
lim (2  h) =2
h 0
This is the gradient (or rate of change) at point A.
In general, for f(x)=x2, the gradient at any x is m=2x.
Justification (not necessary to learn, just for information).
We apply the same technique for any point A(x,x2) and a
neighboring point B with
x-coordinate = x+h
y-coordinate =(x+h)2
Then
Δy
(x  h) 2  x 2 x 2  2xh  h 2  x 2 2xh  h 2

=
=
= 2x  h
Δx
h
h
h
and the gradient (or rate of change) at any x is
lim (2x  h) = 2x
h 0
We write
f (x)  2x
Thus, for example, the gradient at x=3: f (3) =6
The new function f (x) , which is derived from f(x), is called the
derivative of f. Thus
f (x)  DERIVATIVE = RATE OF CHANGE = GRADIENT at x .
8
TOPIC 5: CALCULUS
5.2
Christos Nikolaidis
DERIVATIVES OF KNOWN FUNCTIONS - RULES
The derivative of a function f(x) is a new function denoted by f (x) .
As explained in the preceding section, f (x) indicates the rate of
change, or otherwise the gradient of f(x) at any particular point x.
We have seen that the derivative of the function f(x)=x2 is
f (x) =2x
We can similarly show that for f(x)=x3 the derivative is
f (x) =3x2.
In general, the derivative of the power function f(x)=xn is
f (x) = nxn-1
Look at the table below
f(x)= xn
f (x) =nxn-1
x10
10x9
x4
4x3
x3
3x2
x2
2x
x
1
The derivatives of the most common functions are shown below
f(x)
f (x)
xn
nxn-1
sinx
cosx
cosx
-sinx
ex
ex
1
x
1
lnx
x
2 x
c (constant)
0
9
TOPIC 5: CALCULUS
Christos Nikolaidis
Let us especially elaborate on the power formula (xn)΄ = nxn-1
This formula also applies for –tive values of n:
f(x)= xn
f (x) =nxn-1
x-10
-10x-11
x-3
-3x-4
x-2
-2x-3
x-1
-x-2
It also applies for rational values of n:
f(x)= xn
f (x) =nxn-1
x6.4
6.4x5.4
3 1/2
x
2
5 2/3
x
3
1 -1/2
x
2
x3/2
x5/3
x1/2
EXAMPLE 1

2
 1 
Show that (a)  2   3
x
x 
(b)
 x   21x
Solution
1
2
(a) 2  x  2 , so the derivative is -2x-3 = 3
x
x
1
1
1
(b) x  x 1/2 , so the derivative is
x-1/2 =
=
1/2
2
2x
2 x
EXAMPLE 2
Let f(x)=x7. Find
(a)
f(0), f(1), f(2)
(b)
f (x)
(c)
f (0) , f (1) , f (2)
(d)
the rate of change of f(x) at x=2
(e)
the gradient of f(x) at x=2
10
TOPIC 5: CALCULUS
Christos Nikolaidis
Solution
(a)
f(0)=0,
f(1)=1,
f(2)=128
(b)
f (x) = 7x6
(c)
f (0) =0,
f (1 ) =7.16=7,
f (2) =7.26=448
(d)
it is f (2 ) = 448
(e)
it is f (2) = 448
 NOTATION
If y=f(x), the derivative is denoted by the following symbols
y΄
or
or
f (x)
dy
or
dx
d
f(x)
dx
The derivative at some specific value of x, say x=2, is denoted by
dy
dx x  2
or
f (2 )
For example, if y=f(x)=x3, we can write
y ΄=3x2
or
dy
(x3)΄ = 3x2 or
dx
=3x2
or
d 3
x =3x2
dx
Moreover,
f ΄(2)=12
or
dy
=12
dx x  2
The procedure of finding the derivative is called differentiation.
Notice that the GDC does provide f (x) but it gives the derivative at
a specific point, for example f ΄(2) (check!)
Finally, the notation
dy
is more indicative when we use other
dx
variables.
For example,
ds
 2t .
dt
dP 1
dQ
If P=lnQ, then
 eP .
 . Also Q=eP, so that
dQ Q
dP
if s=t2, then
11
TOPIC 5: CALCULUS
Christos Nikolaidis
 RULES OF DIFFERENTIATION
(f-g)΄ = f ΄ - g΄
(f+g)΄ = f ΄ + g΄
Rule (1):
EXAMPLE 3
For
f(x) = x5+x3 ,
For
g(x)= x7-ex+sinx-x+5,
Rule (2):
f (x) = 5x4+3x2
g΄(x) = 7x6-ex+cosx-1
(a=constant number)
(af)΄ = af ΄
EXAMPLE 4
For
f(x) = 3sinx,
f (x) = 3cosx
For
g(x) = 7ex,
g΄(x) = 7ex
For
h(x) = 5x3,
h΄(x) = 5(3x2)=15x2
Let us combine Rules (1) and (2):
(af + bg)΄ = af ΄+ bg΄
EXAMPLE 5
For
f(x) = 2x3-3x2+7x+5,
For
g(x) = 5x7+3lnx-7cosx,
f (x) = 6x2-6x+7
3
g΄(x) = 35x6+ +7sinx
x
NOTICE:
The differentiation rules above may also be expressed as follows
d
d
d
[f(x) ±g(x)] =
f(x) ±
g(x)
dx
dx
dx
d
d
[af(x)] =a
f(x)
dx
dx
d
d
d
[af(x)  bg(x)] = a
f(x) + b
g(x)
dx
dx
dx
12
TOPIC 5: CALCULUS
Christos Nikolaidis
Rule (3): (product rule)
(f .g)΄ = f ΄.g + f .g΄
Be careful !!!
If f(x) = x5sinx
then f (x) is not (5x4)(cosx)
We must follow the product rule above. That is
f (x) =(x5)΄ sinx + x5 (sinx)΄ = 5x4sinx + x5cosx
EXAMPLE 6
For
f(x) = xlnx,
f (x) = (x)΄lnx + x(lnx)΄ = 1lnx + x
For
g(x) = x2ex,
g΄(x) = 2xex+x2ex
For
h(x) = 2x3cosx,
h΄(x) = 6x2cosx - 2x3sinx
1
= lnx +1
x
Rule (4): (quotient rule)

f   g - f  g
f 
  =
 g
g2
 
EXAMPLE 7
(x 3 )sinx  x 3 (sinx) 3x 2 sinx  x 3 cosx
=
sin 2 x
(sinx) 2
For
x3
f(x) =
,
sinx
For
g(x) =
3x + 5
,
4x + 1
g΄(x) =
3(4x  1)  4(3x + 5)
17

2
(4x  1)
(4x  1)2
For
h(x) =
x 3  5x
,
x2  1
h΄(x) =
(3x 2 - 5)(x 2  1)- (x 3 - 5x)2x
(x 2  1)2
f (x) =

3x 4 + 3x 2 - 5x 2 - 5 - 2x 4 + 10x 2
(x 2  1)2

x 4 +8x 2 - 5
(x 2  1)2
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TOPIC 5: CALCULUS
Christos Nikolaidis
Sometimes, we can avoid the quotient rule. Look at the following
EXAMPLE 8
For
f(x) =
x 3  2x  1
x
method A: The quotient rule gives
f (x) =
(3x 2 - 2)x  (x 3 - 2x  1)1 2x 3  1
1
=
=2x- 2
2
2
x
x
x
method B: we may modify f(x) by splitting into three fractions
f(x) =
x 3 2x 1
+ = x2-2+x-1 , so that
x
x
x
f (x) = 2x-x-2 =2x-
1
x2
 HIGHER DERIVATIVES
We can continue differentiating the 1st derivative f (x) and thus
find the 2nd derivative f (x) , the 3rd derivative f (x) and so on.
The nth derivative is also denoted by f (n) (x) .
EXAMPLE 9

For f(x) = x5,
f (x) = 5x4
f (x) = 20x3
f ΄΄΄(x) = 60x2

For g(x) = sinx
g΄(x) = cosx
g΄΄(x) = -sinx
g΄΄΄(x) = -cosx

For h(x) = ex,
h΄(x) = ex
h΄΄(x) = ex
h΄΄΄(x) = ex
Clearly h (4) (x) = e x and in general h (n) (x) = e x
Alternative notation:
f ΄΄(x) can also be written as
f ΄΄΄(x) can also be written as
14
d2y
dx 2
d3y
dx
3
or
d2
f(x)
dx 2
or
d3
f(x)
dx 3
TOPIC 5: CALCULUS
Christos Nikolaidis
 SOME EXTRA FUNCTIONS (only for HL)
The derivatives of any exponential and any logarithmic function:
f(x)
f (x)
ax
ax .lna
1
xlna
logax
Notice: for a=e we obtain the particular cases, (ex)΄=ex and (lnx)΄=
EXAMPLE 10
For
f(x) = 3x,
f (x) = 3xln3
For
g(x) = 7log5x,
g΄(x) =
7
xln5
The derivatives of the following trigonometric functions:
f(x)
f (x)
tanx
1
 sec2 x
2
cos x
secx
secxtanx
cscx
-cscxcotx
cotx
-csc2x
The derivatives of the inverse trigonometric functions:
f (x)
1
f(x)
arcsinx

arccosx
arctanx
15
1- x 2
1
1- x 2
1
1  x2
1
x
TOPIC 5: CALCULUS
5.3
Christos Nikolaidis
TANGENT LINE - NORMAL LINE AT SOME POINT x0
Remember: A straight line with

gradient m

passing through point (x0,y0)
has equation
y- y0 = m(x- x0)
For example, the line passing through A(1,2) with gradient m=3
has equation
y- 2 = 3(x- 1)
Consider a function y=f(x) and some point x0. Then we also know
y0=f(x0)
y0
x0
We know that the gradient of the graph at (x0,y0) is mT=f ΄(x0).
We define:
TANGENT LINE at x0:
normal line
the line with gradient mT
which passes through (x0,y0)
y0
NORMAL LINE at x0:
the perpendicular line to the
tangent at (x0,y0).
Its gradient is mN= 
1
mT
x0
The point (x0,y0) is also known as point of contact.
16
tangent line
TOPIC 5: CALCULUS
Christos Nikolaidis
 METHODOLOGY
y=f(x)
Given
and some point x=x0
the point of contact (x0,y0) since y0=f(x0)
We find
f (x)
mT=f ΄(x0) and so mN= 
1
mT
The equations of the two lines are
TANGENT LINE
NORMAL LINE
y-y0=mT(x-x0)
y-y0= mN(x-x0)
EXAMPLE 1
Consider the function
f(x)=x2
Find the equations of the tangent line and the normal line at x=3.
Solution
The point of contact is (3,9)
(since f(3)=9)
The gradient function is f (x) =2x. Thus
mT = f (3) =6
and
The tangent line is
y-9=6(x-3)
The normal line is
y-9= 
m N= 
1
6
1
(x-3)
6
After performing the necessary operations we obtain the final
forms
Tangent line:
y-9=6(x-3)  y  9  6x  18  y=6x-9
Normal line:
y-9= 
1
1
1
19
1
(x-3)  y  9  - x   y=  x+
6
2
6
2
6
17
TOPIC 5: CALCULUS
Christos Nikolaidis
NOTICE
An alternative way to obtain the tangent and the normal lines is to
use the formulas:
TANGENT LINE
NORMAL LINE
y  mT x  c
y= m N x+c
The point (x0,y0) helps us to find the constant c.
In the previous example, since m=6 and the point is (3,9):

Tangent line: y=6x+c
At (3,9)

18+c=9  c=-9,
Normal line:
At (3,9) 
y= 
thus
y=6x-9
thus
y= 
1
x+c
6
1
19
1
3+c=9  c=9+  c=
6
2
2
1
19
x+
2
6
EXAMPLE 2
Consider the function
f(x)=5x3-2x+1
Find the tangent lines to the curve which are parallel to the line
L: y=13x+8
Solution
A tangent line parallel to L must have gradient mT=13.
But f (x) = 15x2-2, so
15x2-2=13  15x2=15  x2=1  x=1 or x=-1
Hence, we will have two parallel lines, at the points x=1 and x=-1,
with gradient m=13. The points of contact are (1,4) and (-1,-2)

At (1,4),
y-4=13(x-1)  y-4=13x-13  y=13x-9

At (-1,-2)
y+2=13(x+1)  y+2=13x+13  y=13x+11
18
TOPIC 5: CALCULUS
Christos Nikolaidis
NOTICE
Suppose that the gradient at (x0,y0) is mT=0. Then
The tangent line is a horizontal line with equation y=y0
The normal line is a vertical line1 with equation x=x0
EXAMPLE 3
Consider the function f(x) = x 2 - 4x + 5
Find the equations of the tangent line and the normal line at x=2.
Solution
It is f (x) =2x-4
At x=2, y=1, thus the point of contact is (2,1)
mT = 0
(mN is not defined)
Tangent line: the horizontal line y=1
Normal line: the vertical line x=2 (look at the graph above!)
For trickier questions, let’s have in mind the following observation.
At the point of contact between f(x) and a tangent line y=mx+c
functions are equal:
f(x)=y
derivatives are equal:
f (x) =m
1 A vertical line has no gradient. If it passes through (x
19
0,y0), it has equation x=x0.
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 4
The line y  mx - 3 is tangent to the curve f(x)=x4-x. Find m
Solution
Αt the point of contact x:
functions are equal:
x4-x  mx - 3
derivatives are equal:
4x3-1  m
Hence,
x4-x  (4x 3 - 1)x - 3  x 4 - x  4x 4 - x - 3
 3x 4  3
 x 4  1  x  1
If x  1
then m=3
If x  1
then m=-5
EXAMPLE 5
Consider the function f(x)=x4-x. Find the tangent lines passing
through the point (0,-3) [notice that the point is not on the curve]
Method A:
A line passing through the point (0,-3) has the form
i.e
y  3  m(x - 0)
y  mx - 3
This is in fact the example 4 above. We found two solutions:
If x  1
then m=3
and the tangent line is y  3x - 3
If x  1 then m=-5
and the tangent line is y  -5x - 3
Method B:
We find the tangent line at any point (a,f(a), i.e. (a,a4-a).
f (x) =4x3-1, so mT = 4a3-1. Therefore,
y-(a4-a) = (4a3-1)(x-a)
 y = (4a3-1)x -3a4
It passes through (0,-3):
-3a4 = -3  a4 =1  a=±1.
For a =1 we obtain the equation y  3x - 3 .
For a =-1 we obtain the equation y  -5x - 3 .
20
TOPIC 5: CALCULUS
5.4
Christos Nikolaidis
THE CHAIN RULE
Look at the differentiation table again
f(x)
f (x)
sinx
cosx
cosx
-sinx
ex
ex
1
x
1
lnx
x
2 x
x3
3x2
In simple words the chain rule says:
If we replace x by another function u(x) then the derivative is as
above but we also multiply the result by u΄(x).
Let us repeat the table above where x is replaced by u(x):
f(x)
f (x)
sin u(x)
cos u(x) . u(x)
cos u(x)
-sin u(x) . u(x)
eu(x)
eu(x) . u(x)
1 .
u(x)
u(x)
1
. u(x)
2 u(x)
ln u(x)
u(x)
u(x)3
3u(x)2 . u(x)
EXAMPLE 1
f(x)
= sin(2x2+3)
[Here u=2x2+3]
f (x) = cos(2x2+3) (2x2+3)΄
= 4x cos(2x2+3)
21
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 2
f(x)
= e5x+3
[Here u=5x+3]
f (x) = e5x+3 5
= 5e5x+3
EXAMPLE 3
f(x)
= esinx
[Here u=sinx]
f (x) = esinxcosx
EXAMPLE 4
f(x)
=ln (x2+4)
f (x) =
[Here u= x2+4]
1
2x
x 4
2
=
2x
x 4
=
3x 2  5x  2
2
EXAMPLE 5
f(x)
f (x) =
=
1
2
2 3x  5x  2
[Here u=3x2+5x+2]
(6x+5)
6x  5
2 3x 2  5x  2
EXAMPLE 6
f(x)
=
f (x) =
[Here u=cosx]
cosx
-sinx
2 cosx
22
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 7
Consider the function f(x) = (2x2+3)2
If we first expand: f(x)=4x4+12x2+9 then f (x) = 16x3+24x
If we follow the chain rule [with u=2x2+3]
f (x) = 2(2x2+3)(4x) =8x(2x2+3) = 16x3+24x
But what about
f(x) = (2x2+3)10 ?
Do not attempt to expand, it will be laborious! The chain rule gives
f (x) = 10 (2x2+3)9 (4x) = 40x (2x2+3)9
NOTICE
In many examples u=ax+b (linear) so that u΄=a. Hence,
we simply differentiate for u and multiply the result by a.
EXAMPLE 8
Let us consider all the usual functions with u=3x+7.
f(x)
f (x)
sin(3x+7)
3cos(3x+7)
cos(3x+7)
-3sin(3x+7)
e3x+7
3e3x+7
3
3x  7
3
ln(3x+7)
3x  7
2 3x  7
(3x+7)5
15(3x+7)4
23
TOPIC 5: CALCULUS
Christos Nikolaidis
Next, instead of x we have u=5x.
Perhaps
the
most
f(x)
f (x)
sin(5x)
5cos(5x)
cos(5x)
-5sin(5x)
e5x
5e5x
ln(5x)
5
1

5x x
confusing
case
of
the
chain
rule
is
the
differentiation of the function sinnx (or cosnx). Remember that
sinnx
means
(sinx)n
[so that u=sinx]
EXAMPLE 9
For f(x) = sin3x,
f (x) = 3sin2x cosx
For f(x) = sin2x,
f (x) = 2sinx cosx
For f(x) = cos5x,
1
For f(x) =
 (sinx)-1,
sinx
1
For f(x) =
 (cosx)-2,
cos 2 x
f (x) = -5cos4x sinx
cosx
sin 2 x
sinx
f (x) = -2(cosx)-3(-sinx) =
cos3 x
f (x) = -(sinx)-2 cosx = 
NOTICE
In fact, the chain rule is the rule of differentiation for the
composition of two functions
(f g)(x) = f(g(x))
It says that
f(g(x)   f (g(x))g(x)
I admit that this definition is not so “elegant”! The best way to
learn the chain rule is to practice with a great deal of examples.
24
TOPIC 5: CALCULUS
Christos Nikolaidis
It is possible to have a “double chain”, that is a chain inside
another chain! It is in fact the derivative of the composition of
three functions. The following example is indicative.
EXAMPLE 10
a) Let f(x) = ln(sin(3x+1))
f (x) =
1
[sin(3x+1)]΄
sin(3x  1)
[u =sin(3x+1)]
=
1
[3cos(3x+1)]
sin(3x  1)
[v = 3x+1]
=3
cos(3x  1)
sin(3x  1)
b) Let f(x) = 3sin5(x2+1)
f (x) =15 sin4(x2+1)[sin(x2+1)]΄
= 15 sin4(x2+1)cos(x2+1)(2x)
[u =sin(x2+1)]
[v = x2+1]
= 30xsin 4 (x 2  1)cos(x 2  1)
c) Let f(x) = e sin(3x)
f (x) = e sin(3x) [sin(3x)]΄
= e sin(3x) [3cos(3x)]
[u =sin(3x)]
[v = 3x]
= 3 e sin(3x) cos(3x)
d) Let f(x) =
sin 2 x  sin2x
f (x) =
=
=
1
2
2 sin x  sin2x
1
2 sin 2 x  sin2x
sinxcosx  cos2x
sin 2 x  sin2x
25
(sin 2 x  sin2x)
(2sinxcosx  2cos2x)
TOPIC 5: CALCULUS
Christos Nikolaidis
We may also have a combination of all the rules we have seen.
EXAMPLE 11
Find the derivative of f(x) = e2xsin3x.
Solution
We have a product of two functions e2x and sin3x, but for each
function we must apply the chain rule
f (x) = (e 2x )(sin3x) + (e 2x )(sin3x)
= 2e 2x sin3x + 3e 2x cos3x
EXAMPLE 12
2
Find the derivative of f(x) = e x sinx .
Solution
This is an example of chain rule where u = x 2 sinx is a product.
2
f (x) = e x sinx  x 2 sinx 
= e x sinx  2xsinx + x 2 cosx 
2
Finally, let us see some slightly more abstract examples:
EXAMPLE 13
Differentiate the following functions (of the first column):
y = sinf(x)
y = f(sinx)
y = f(x)5
y = f(x 5 )
dy
= cosf(x)  f (x)
dx
dy
= f (sinx)  cosx
dx
dy
= 5f(x) 4  f (x)
dx
dy
= f (x 5 )  5x 4
dx
26
TOPIC 5: CALCULUS
Christos Nikolaidis
 AN ALTERNATIVE WAY TO LOOK AT THE CHAIN RULE
Let y depend on u, and u depend on x:
We have the chain
y
u
x
The chain rule says
dy
dx
=
dy du
du dx
[Notice: it is easy to remember this formula as it looks like a
simplification of fractions!!!]
Look at again
y = (2x2+3)10
Then
y = u10
where
u=2x2+3
The chain rule gives
dy
=
dx
dy du
du dx
= 10u9 (4x)
= 10(2x2+3)9(4x)
[replace back u=2x2+3]
= 40x(2x2+3)9
[as found earlier]
EXAMPLE 14
Let y= esinx.
Then y =eu were u=sinx.
dy
dx
=
dy du
du dx
= eu cosx = esinx cosx
[this is what we obtain if we apply the chain rule as usual]
27
TOPIC 5: CALCULUS
Christos Nikolaidis
The following example also justifies the term “chain”!!!
EXAMPLE 15 (Mainly for HL)
Let P = Q3 and Q=lnR. Find
We have the chain
dP
in terms of R
dR
P
Q
R
(as P depends on Q and Q depends on R)
The chain rule gives
dP
dR
dP . dQ
dQ dR
1
= 3Q2 .
R
=
= 3(lnR)2
1
R
28
TOPIC 5: CALCULUS
5.5
Christos Nikolaidis
MONOTONY – MAX, MIN
 INCREASING – DECREASING FUNCTIONS (MONOTONY)
Consider the following graph
y=f(x)
a
c1
c2
c3
c4
c5
b
Let us make some observations:

The domain of the graph is the interval [a,b]
The points x=a and x=b are called endpoints

We say that we have a local max (or just max) at points:
x=c1, x=c3, x=b
(can you explain why?)

We say that we have a local min (or just min) an points:
x=a, x=c2, x=c4
(can you explain why?)
All these points (max and min) are called turning points or
extreme values

Notice that x=c5 is not a turning point (neither max nor min),
as near f(c5) you can find smaller as well as larger values.

The function is increasing (goes up) in the interval (a,c1)

The function is decreasing (goes down) in the interval (c1,c2)

The function is increasing (goes up) in the interval (c2,c3)

The function is decreasing (goes down) in the interval (c3,c4)

The function is increasing (goes up) in the interval (c4,b)
29
TOPIC 5: CALCULUS
Christos Nikolaidis
Remember though that
A +tive gradient means that the function is increasing (goes up)
A -tive gradient means that the function is decreasing (goes down).
But we know that
derivative
=
gradient
In other words
If f  (x)  0
then
f is increasing (
)
If f  (x)  0
then
f is decreasing (
)
Notice: The increasing or decreasing behavior of a function is also
known as monotony!
 TURNING POINTS: MAX - MIN
How can we find the turning points (max or min) of a function?
First, the end points are extreme values (see a and b above).
As far as the interior points is concerned, observe that the gradient
at any turning point is 0 (the tangent lines at those values are
horizontal!).
PROPOSITION:
If f(x) has a turning point (max or min) at some interior point c
and f ΄(c) exists, then
f ΄(c)=0
Notice that f ΄(x)=0 at c1, c2, c3, c4 and c5
We have a local max at c1, c3 and a local min at c2, c4.
However, c5 is not a turning point.
Hence the inverse proposition is not true.
30
TOPIC 5: CALCULUS
Christos Nikolaidis
Therefore, apart from the endpoints, the possible turning points
(max/min) are the following

points x where f (x) =0 (called stationary points)

points where f (x) does not exist
All these points are also called critical points.
Here we only deal with stationary points (points where f (x) =0).
To verify whether such a stationary point x=c is a turning point
(max or min) we perform the following test
FIRST DERIVATIVE TEST for c
Check the sign of f (x) to verify if the function is increasing or
decreasing just before and after c:
x
x
c
f ΄(x)
+

Conclusion for f
c
f ΄(x)


Conclusion for f
max
+

min
If the sign does not change, we have neither a max nor a min.
 METHODOLOGY
Given
y=f(x)
Step 1
we find f (x)
Step 2
we solve f (x) =0 (say that roots are a,b,c)
Step 3
we construct a table as follows to perform the
first derivative test
x
f ΄(x)
Conclusion for f
a
+

b

max
31
c
+

min
+

nothing
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 1
Consider
f(x)=
We find
1 3
x -2x2+3x+5
3
f (x) =x2-4x+3
We solve
x2-4x+3=0
The solutions are x=1 and x=3
We construct the table
x
f ΄(x)=x2-4x+3
Conclusion for f
1
+

3

max
+

min
Therefore,
we have a max at x=1 [and the max value of f is f(1)=6.33]
we have a min at x=3 [and the min value of f is f(3)=5]
An alternative way to check if a stationary point is a max or a min
is the following:
SECOND DERIVATIVE TEST for c
Find f (x)
(if it exists!)
If f (c) > 0 then c is a min
If f (c) < 0 then c is a max
If f (c) =0 we don’t get an answer. We go back to the first
derivative test.
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TOPIC 5: CALCULUS
Christos Nikolaidis
Let us use the same example as above
EXAMPLE 2
Consider
f(x)=
We found
1 3
x -2x2+3x+5
3
f (x) =x2-4x+3
and the stationary points x=1 and x=3
We find
f (x) =2x-4
For x=1, f (1 ) =-1<0, so we have a max at x=1
For x=3, f (3 ) = 2>0, so we have a min at x=3
EXAMPLE 3
Consider
f(x)= (x-1)4
We find
f (x) =4(x-1)3
There is only one stationary point at x=1.
f (x) = 12(x-1)2
For x=1, f (1 ) = 0 (neither positive nor negative).
Thus, we cannot conclude if it is a max or a min.
The table of signs gives
x
1
f ΄(x)=4(x-1)3

Conclusion for f(x)
+

min
Therefore, we have a min at x=1.
33
TOPIC 5: CALCULUS
5.6
Christos Nikolaidis
CONCAVITY - POINTS OF INFLEXION
 CONCAVITY
Consider again the graph of the preceding section
y=f(x)
d1
d2
d3
d4
d5
d6
d7
Our concern now is different! It is to investigate the intervals
where the curve
looks like () : we say that the function is concave up
looks like () : we say that the function is concave down2
We observe that:

The function is concave down () in the interval (d1,d2)

The function is concave up () in the interval (d2,d3)

The function is concave down () in the interval (d3,d4)

The function is concave up () in the interval (d4,d5)

The function is concave down () in the interval (d5,d6)

The function is concave up () in the interval (d6,d7)

The concavity changes at the points x=d2, d3, d4, d5, d6, d7
These points are called points of inflexion or inflexion points
It is easy to verify the concavity by using the second derivative
2
If f ( x ) > 0
then
f is concave up (  )
If f ( x ) < 0
then
f is concave down (  )
To be more formal, a function is concave up/down if the tangent line at each
point is under/above the curve.
34
TOPIC 5: CALCULUS
Christos Nikolaidis
Short explanation (mainly for HL)
Look at the curve of the following concave up function f(x)
The gradient is -tive in the beginning, it is “less” -tive as we move
forward, it sometimes becomes 0 and then becomes +tive and
“more” +tive as we move forward. In other words the gradient
increases. That is, the function of the gradient f ( x ) is increasing.
But we know that the derivative of an increasing function is +tive.
Hence, the derivative of f ( x ) , that is f ( x ) is +tive!
Similarly, if the function f(x) is concave down, the second derivative
must be -tive!
 POINTS OF INFLEXION
How can we find the points of inflexion?
Since the concavity changes at such a point, the sign of f ( x )
changes from + to – or vice-versa. Therefore, the second derivative
at any point of inflexion must be 0.
PROPOSITION:
If f(x) has a point of inflexion at some point d and f ( d ) exists,
then
f ( d ) =0
Notice again that the equation f ( x ) =0 gives us the possible points
of inflexion. To verify if x=d is indeed a point of inflexion we must
check the sign of f ( x ) just before and after that point.
35
TOPIC 5: CALCULUS
Christos Nikolaidis
 METHODOLOGY
Given
y=f(x)
Step 1
we find f ( x ) and f ( x )
Step 2
we solve f ( x ) =0 (say that roots are a,b,c)
Step 3
we construct a table as follows
x
a
f (x )
Conclusion for f
b
c
+
-
+
+




i.p.
i.p.
nothing
EXAMPLE 1
Consider again
f(x)=
We find
1 3
x -2x2+3x+5
3
f ( x ) =x2-4x+3
f ( x ) =2x- 4
We solve
2x- 4=0
The solution is x=2
We construct the table
x
2
f ( x ) =2x-4
Conclusion for f
-
+


i.p.
Therefore, x=2 is an inflexion point.
Let us summarize the information we have for our example
f(x)=
1 3
x -2x2+3x+5
3
in order to sketch the graph of this function.
36
TOPIC 5: CALCULUS
Christos Nikolaidis
First, it helps to know the y-intercept and the x-intercepts, that is
the roots of f(x) (if possible):
y-intercept: for x=0, y=f(0)=5
x-intercepts: we must solve f(x)=0 (only by GDC as the degree is 3)
We consider a summary table containing the solutions of both
f ( x ) =0 and f ( x ) =0:
x
1
2
3
f (x )
+
-
-
+
f (x )




+


+


Conclusion for f
max
i.p.
min
The following table of values will also help
x
0
1
2
3
f(x)
5
6.33
5.66
5
y
7
max
i.p.
6
5
min
4
3
2
1
x
-3
-2
-1
1
2
3
4
5
6
7
-1
EXAMPLE 2
Consider the function
f(x)= x e x
Find possible maximum, minimum values and points of inflexion.
37
TOPIC 5: CALCULUS
Christos Nikolaidis
Solution
We have
f ( x ) = x e x + e x
Stationary points:
x e x + e x =0  e x (x+1)=0  x=-1
We use table:
x
-1
f ( x )
-
Conclusion for f
+
min
Furthermore,
f ( x ) = x e x + e x + e x = x e x +2 e x
Then
x e x +2 e x =0  e x (x+2)=0  x=-2
We use table:
x
-2
f (x )
Conclusion for f
-
+


i.p.
The graph of the function is shown below
Notice: Look at the graph of this function at your GDC to confirm
that x=-1 gives a min and observe that at x=-2 there is a point of
inflexion.
38
TOPIC 5: CALCULUS
5.7
Christos Nikolaidis
OPTIMIZATION
In problems of optimization, we have to construct a function in
terms of some variable x, and then we use derivatives to find the
“optimum” solution, that is the maximum or the minimum value
of the function.
EXAMPLE 1
Among all the rectangles of perimeter 20, find the one of the
maximum area.
Discussion
A rectangle of perimeter 20 may have dimensions
1×9
2×8
3×7
4×6
etc
21
24
etc
The corresponding areas are
9
16
Which is the one of the maximum area?
Solution
y
x
Let x be one of the sides (this will be our main variable).
If the other side is y, then
Perimeter = 20  2x +2y =20  y = 10-x
The function of optimisation is
Area:
A = xy = x(10-x)=10x-x2
dA
 10  2x
dx
dA
Stationary points:
= 0  10  2x = 0  x =5
dx
We find
The 2nd derivative test is easier here: A΄΄ = -5.
At x=5 A΄΄ < 0, thus we have a maximum value there.
39
TOPIC 5: CALCULUS
Christos Nikolaidis
Therefore, the rectangle of maximum area is the square (that is
when x = 5), and the maximum area is Amax = 25.
Let’s reverse the role of the perimeter and the area. Next we know
the area of the rectangle and we are looking for the minimum
perimeter.
EXAMPLE 2
Among all the rectangles of area 25, find the one of the minimum
perimeter.
Solution
y
x
Again, let x be one of the sides (this will be our main variable).
If the other side is y, then
Area = 25  xy=25  y =
25
x
The function of optimisation is
Perimeter:
We find
P = 2x+2y = 2x+
50
x
dP
50
= 2 2
dx
x
Stationary points:
dP
50
= 0  2  2 = 0  x 2 = 25  x = 5
dx
x
The 2nd derivative test gives: P΄΄ =
100
.
x3
For x=5, P΄΄ > 0 , thus we have a minimum value there.
Therefore, the rectangle of minimum perimeter is the square, (that
is when x = 5), and the minimum perimeter is Pmin = 25.
40
TOPIC 5: CALCULUS
Christos Nikolaidis
Sometimes, there is a different “cost” for each side.
EXAMPLE 3
We want to construct a rectangle fence for an area of 24m2, but
the cost for the material of the front side is 10$ per meter while
the cost for the material of the other 3 sides is 5$ per meter. Find
the cheapest solution!
Solution
y
x
Let x be the front side (this will be our main variable).
If the other side is y, then
Area = 24  xy=24  y =
24
x
The function of optimisation is
Cost:
We find
C = 10x+5x+2(5y) = 15x+10y = 15x 
240
x
dC
240
= 15  2
dx
x
Stationary points:
dC
240
=0  15  2 = 0  x 2 = 16  x = 4
dx
x
The 2nd derivative test gives: C΄΄ =
480
.
x3
For x=4, C΄΄ > 0 , thus we have a minimum value there.
Therefore,
the
best
rectangle
has
dimensions
4×6
and
the
minimum cost is Cmin = 120$
This rationale applies to 3D shapes as well. For example, they may
give us a rectangular prism or a cylinder of a given volume and we
are looking for the optimum surface area, or vice-versa.
41
TOPIC 5: CALCULUS
Christos Nikolaidis
Let’s see an example of a shape determined by the boundaries of a
given function!
EXAMPLE 4
Consider the region enclosed by y  9  x 2 and x-axis.
Find the rectangle of largest area inscribed within that region.
y
10
5
x
-4
-3
-2
-1
1
2a
3
4
Discussion
There are two extreme cases:

the height of the rectangle is 0, the width is 6. The area is 0.

the height of the rectangle is 9, the width is 0. The area is 0.
Somewhere in between there is a rectangle of maximum area.
Solution
Key point: Let a be the x-coordinate of the bottom right corner.
Then
Width = 2a
Height = y  9  a 2
[it is f(a) !]
Thus, the function of optimisation is
Area:
A = 2a(9-a2) = 18a-2a3
dA
 18  6a 2
da
dA
Stationary points:
= 0  18  6a 2 = 0  a =
dx
We find
3
The 2nd derivative test gives: A΄΄ = -12a.
At a= 3 , A΄΄ < 0, thus we have a maximum value there.
Therefore, the rectangle of maximum area has dimensions 2 3 ×6
and the maximum area is Amax = 12 3 .
42
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 5
A
5km
3km
4km
C
B
B
A swimmer is at point A inside the sea, 3km away from the beach.
She wants to go to point B at the beach, which is 5 km away.
When she swims she covers 1 km in 8 minutes
When she runs she covers 1km in 5 minutes. Find
(a)
the time she spends when she swims directly to B;
(b)
the time she spends when she swims to C and then runs to B;
(c)
the minimum time she can achieve if she swims first to some
point D between B and C and then runs to B
Solution
(a) T=TAB= 58 = 40 min
(b) T= TAC + TCB = 38 + 45 = 44 min
(c) Let D be between C and B and CD=x km. Then DB = 4-x km
Also AD2 = AC2 + CD2  AD =
9 x2
Therefore, the function of optimization is the total time
T= TAD + TDB = 8 9  x 2 + 5(4-x) = 8 9  x 2 -5x+20 min
dT
8x

5
dx
9 x2
8x
Stationary points:
 5  0  8x  5 9  x 2
2
9 x
We find
 64 x 2  25(9  x 2 )  39x 2  225  x 
15
 2.4 km
39
We can easily verify that this is a minimum (2nd derivative test).
Thus, the point D is 2.4km from C.
The minimum time is Tmin=38.7 minutes.
43
TOPIC 5: CALCULUS
5.8
Christos Nikolaidis
THE INDEFINITE INTEGRAL  f(x)dx
 THE INDEFINITE INTEGRAL
Consider
F(x)=x2
The derivative is
F (x) = 2x
The reverse problem: If they give us the result
f(x)=2x
can we find a function F(x), such that F (x) = f(x) ?
Of course, one answer is F(x)=x2. We say that F(x)=x2 is an
antiderivative of f(x)=2x. But it is not the only one!
Notice that
(x2)΄ = 2x
(x2+1)΄ = 2x
(x2+2)΄ =2x
(x2+5)΄ =2x
in general
(x2+c)΄ = 2x
for any constant c
Therefore, the functions x2+c are also antiderivatives of f(x)=2x.
We say that x2+c is the indefinite integral of f(x)=2x and we use the
notation
 2xdx = x2+c
Hence,
if
F ΄(x) = f(x)
then  f(x)dx = F(x)+ c
For example,
since (x5)΄=5x4
we obtain
4
 5x dx = x5 + c
4
 x dx =
We also deduce that
44
x5
+ c (why?)
5
TOPIC 5: CALCULUS
Christos Nikolaidis
Therefore, we can easily obtain the following results
f(x)
 f(x)dx
1
x +c
x
x2
x3
x10
x2
+c
2
x3
+c
3
x4
+c
4
x 11
+c
11
In general,
n
 x dx =
x n 1
+ c
n 1
(if n  -1)
Notice also,
 5dx = 5x + c, since (5x)΄=5
 adx = ax + c, since (ax)΄=a (a=constant)
since (sinx)΄=cosx
 cosxdx = sinx + c,
If we remember the derivatives of the basic functions we obtain the
following results:
f(x)
 f(x)dx
xn
x n 1
n 1
a
ax
ex
ex
sinx
-cosx
cosx
sinx
1
x
lnx
+ c
NOTICE: The operation of finding the integral is called integration.
45
TOPIC 5: CALCULUS
Christos Nikolaidis
n
 x dx =
 REMARK FOR
x n 1
+ c
n 1
The same formula applies for negative values of n. For example,
-5
 x dx =
What about 
x-4
1
+c =
+c
-4
- 4x 4
1
1
dx ? We know that 2 =x-2, so
2
x
x
-1
1
x
1
-2
 x 2 dx =  x dx = - 1 +c = - x +c
Also, the same formula applies for rational values of n:
3/5
 x dx =
x 8/5
5 8/5
+c =
x
+ c
8/5
8
What about  x dx ? We know that

x dx =  x 1/2 dx =
x =x1/2 , so
x 3/2
2 3/2
+c =
x
+ c
3/2
3
1
dx =  x -1 dx .
x
1
Only for this particular power we have the formula  dx =lnx+c
x
Notice that this formula does not apply for 
Thus, for example
 f(x)dx
f(x)
1
 x-2
x2
1
 x-3
x3
1
 x-4
4
x
3
2
x2  x 3
x-1
+c
-1
x-2
+c
-2
x-3
c
-3
3 53
x c
5
46
1
c
x
1

c
2x 2
1

c
3x 3
33 5
x c
5

TOPIC 5: CALCULUS
 REMARK FOR 
Christos Nikolaidis
1
dx (only for HL)
x
1
In fact
 x dx = ln|x|+c .
Indeed:
if x>0, then [ln|x|]΄ =[lnx]΄=
1
,
x
if x<0, then [ln|x|]΄ =[ [ln(-x)]΄=
That is why the antiderivative of
1
1
(-1)=
.
-x
x
1
is ln|x|+c.
x
 TWO BASIC RULES OF INTEGRATION
 [f(x)  g(x)]dx =  f(x)dx   g(x)dx
Rule 1:
For example,
4
2
4
2
 (x  x )dx   x dx   x dx 
Rule 2:
x5 x3

c
5
3
 af(x)dx = a  f(x)dx
For example,
4
4
 10x dx  10  x dx  10
x5
 c  2x 5  c
5
In practice, if we have a “long” expression like
 [af(x)  bg(x)  ch(x)  dk(x)]dx
where a, b, c, d are constant coefficients, we keep a, b, c, d and
integrate only f(x), g(x), h(x), k(x).
EXAMPLE 1
2
x
2
x
 [3x  5e  2cosx]dx = 3  x dx +5  e dx -2  cosxdx
x3
)+ 5ex -2sinx + c
3
= 3 x 3 + 5ex -2sinx + c
= 3(
47
TOPIC 5: CALCULUS
Christos Nikolaidis
In fact, the shaded step above is not necessary! We can proceed to
the next step by estimating directly the integrals of x2, ex, cosx
EXAMPLE 2
x5
x4
x3
x2
+8
-5
-7
+2x + c
5
4
3
2
2 5
5
7
=
x +2 x 4 - x3 - x2 + 2x + c
5
3
2
4
3
2
 [2x  8x  5x  7x  2]dx = 2
EXAMPLE 3
2 8 5
-4
-3
-2
 [ x 4  x 3  x 2  2]dx   [2x  8x  5x  2]dx
x-3
x-2
x-1
+8
-5
+2x + c
= 2
-3
-2
-1
2
4 5
= 
 2+
+ 2x + c
3
3x
x
x
 FIND the constant c
Sometimes, we are given an extra condition of the form f(a)=b in
order to find the value of the constant c. We find in fact, the
specific antiderivative which satisfies this condition.
EXAMPLE 4
Let f ΄(x)=6x2- 4x+5. Find f(x) given that f(1)=8.
Clearly, f(x) is the integral of f ΄(x)=6x2- 4x+5. That is,
f(x)=  [6x 2  4x  5]dx = 6
x3
x2
- 4
+5x+c = 2x3-2x2+5x+c
3
2
Next, we have to find the value of c:
f(1)=8,
so
2(1)3-2(1)2+5(1)+c=8
so
5+c=8
so
c=3
Therefore,
f(x) = 2x3-2x2+5x+3
48
TOPIC 5: CALCULUS
Christos Nikolaidis
 TWO MORE BASIC INTEGRALS (only for HL)
In the IB Math HL formula booklet you will also find the formulas
1
1
x
 a  x dx  a arctan a  c
2
2
1
x
 a  x dx  arcsin a  c
2
2
Indeed, the derivative of f(x) =
f (x) 
1
a
1
x
1 
a
2

1
x
arctan is
a
a
1 1
1
1
 2 2
2 
2
a a a x
a  x2
a2
Similarly, the derivative of f(x) = arcsin
1
f (x) 
x
1 
a
2

x
is
a
1 1
1
1


a a a2  x 2
a2  x 2
a2
Later on we will present a different proof of those two formulas.
Let’s find some integrals of this form
EXAMPLE 5
1
 1  x dx  arctanx  c
1
2
1
x
 4  x dx  2 arctan 2  c
5
2
5
x
 13  x dx  13 arctan 13  c
5
2
5
 9  4x dx  4  9
2
4
1
 x2
dx 
52
2x
5
2x
arctan
 c = arctan
c
43
3
6
3
49
TOPIC 5: CALCULUS
Christos Nikolaidis
Similarly
EXAMPLE 6
1
 1  x dx  arcsinx  c
2
1
x
 4  x dx  arcsin 2  c
2
5
x
 13  x dx  5arcsin 13  c
2
5
5
 9  4x dx  2  9
2
4
1
 x2
dx 
5
2x
arcsin
c
2
3
50
TOPIC 5: CALCULUS
5.9
Christos Nikolaidis
INTEGRATION BY SUBSTITUTION
Before we present this method of integration, let us see a simple
case where we have (ax) or (ax+b) instead of x in our function.
 THE LINEAR CASE:  f(ax)dx or  f(ax  b)dx
Consider the integrals
 cos(3x)dx
 cos(3x  5)dx
They look like  cosxdx but instead of x we have u=3x and u=3x+5
respectively.
If we have ax or ax+b instead of x, then at the result
we also write ax+b instead of x,
but we divide the result by a
Hence,
 cos(3x)dx =
sin(3x)
+c
3
 cos(3x  5)dx =
sin(3x + 5)
+c
3
[u=3x]
[u=3x+5]
Indeed, if we differentiate the results
the derivative of
sin(3x)
cos(3x)
is
×3 = cos(3x)
3
3
the derivative of
sin(3x + 5)
is cos(3x+5)
3
Notice that,
 cos(x + 5)dx = sin(x+5) +c
since the coefficient of x is 1.
Indeed,
the derivative of sin(x+5) is cos(x+5).
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TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 1
3x  2
 e dx
-5x  2
e
dx
10x
 e dx
e
x 8
dx
e 3x  2
+c
3
e-5x  2
=
+c
-5
=
=
e 10x
+c
10
= ex+8 +c
 sin(2x  1)dx = 
cos(2x  1)
+c
2
 sin2xdx
cos2x
+c
2
= 
 cos(5 - x)dx =
1
 7x  3 dx
1
 3- 7x dx
1
 x  3 dx
sin(5 - x)
+c = -sin(5-x) +c
-1
=
ln 7x  3
+c
7
=
ln 3- 7x 
+c
-7
= ln(x-3) +c
5
 (3x  5) dx =
(3x  5) 6
+c
3 6
(3x  5)-4
(3x  5)-4
1
-5
=
=
+c
=
+c
(3x

5)
dx
dx
 (3x  5)5

-4  3
-12

3x  5dx =  (3x  5)1/2 dx =
2 (3x  5)3/2
2
+c =
(3x  5)3/2 +c
3
3
9
EXAMPLE 2
Find I   cos2 xdx by using the double angle formula cos2x 2cos2 x  1
Solution
1  cos2x
. Thus
2
1
sin2x 
1
x sin2x
I   (1  cos2x)dx   x 
c
c  
2
2
2 
2
4
If we solve for cos2x we get cos 2 x 
52
TOPIC 5: CALCULUS
Christos Nikolaidis
 SOME DISCUSSION ABOUT THE DIFFERENTIAL dx
(you may skip this page, it is extra information)
The quantity dx is called the differential of x.
It indicates a little change in x. (Compare with Δx=x2-x1)
For a function y=f(x), we know that
dy
Although
dy
dx
dx
 f (x)
is a symbol and not a fraction, it behaves like a
fraction. Besides, it comes from a fraction
Δy
Δx
.
Thus we can solve for dy and write
dy  f (x)dx
This gives a relation between the differentials dy and dx.
In fact, this relation indicates in what extend a little change in x
causes a little change in y.
For example, consider the function y=x2. Then
dy
dx
=2x 
dy=2xdx
We can also solve for dx and write
dx=
dy
2x
.
Similarly, for y=sinx
dy
dx
=cosx

dy=cosxdx
or
dx=
53
dy
cosx
TOPIC 5: CALCULUS
Christos Nikolaidis
 THE METHOD OF SUBSTITUTION (THE SIMPLE CASE)
In fact, we have already seen a simple version of substitution,
where an expression of the form u=ax+b appears in the integral.
For example, we have seen that
I=  cos(3x  5)dx 
1
sin(3x  5)  c
3
Let us use this example to demonstrate analytically the method of
substitution.

Let
u=3x+5

Then
du
du
 3  dx 
dx
3

The integral becomes
I=  cosu
Thus,
I=
du 1
1
  cosudu  sinu  c
3 3
3
1
sin(3x  5)  c
3
If u=ax+b we can directly write down the result as explained. But
for more difficult cases we follow this process.
In general, our target is to eradicate x and dx from the original
integral and obtain a simpler integral in terms of u and du.
METHODOLOGY:
We have to imagine an appropriate substitution u=g(x)
[criterion: g (x) must also exist in the integral as a factor]

Let u=g(x)

Then

Express the initial integral in terms of u and du

Calculate the new integral

Replace u= g(x) back in the result
du
du
= g (x)  dx=
dx
g (x)
54
TOPIC 5: CALCULUS
Christos Nikolaidis
Consider
I=  3x 2 (x 3  5)7 dx
We let u=x3+5 [since the derivative 3x2 exists inside the integral]
Let
u=x3+5
Then
du
du
=3x2  dx=
dx
3x 2
The integral obtains the convenient form
I=  3x 2 u7
du
=  u7 du
2
3x
Hence,
u8
c
8
Finally, we replace back u=x3+5 to get
I=
I=
(x 3  5)8
c
8
Notice that
I=  x 2 (x 3  5)7 dx
can be treated in the same way. The derivative of u=x3+5 is 3x2.
We are happy that x2 exists inside the integral (we don’t mind for
the constant 3). Thus,
let
u=x3+5
then
du
du
=3x2  dx=
dx
3x 2
The result is
I=  x 2 u7
(x 3  5)8
1 u8
du 1 7
=
=
=
u
du

c
c
3 8
24
3x 2 3 
Sometimes, the substitution is given as a hint! See next example.
55
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 3
I=  x x 2  3dx by using the substitution u=x2+3.
Find
Solution
Let
u=x2+3, then
du
du
=2x  dx=
dx
2x
Thus,
I=  x u
3
3
du 1
1 2 2
1 3
1
=  udu =
u +c= u 2 +c= (x 2  3) 2 +c
2x 2
2 3
3
3
EXAMPLE 4
I= 
Find
2x  cosx
dx .
x 2  sinx
Solution
Notice that the derivative of the denominator is the numerator.
u= x 2  sinx , then
Let
du
du
=2x+cosx dx=
dx
2x  cosx
Thus,
I= 
2x  cosx
du
1
=  du =lnu+c=ln (x 2  sinx)  c
u
2x  cosx
u
EXAMPLE 5
I 1= 
Find
(lnx) 2
dx ,
x
I 2= 
1
dx ,
x(lnx) 2
I 3= 
Solution
For all three of them we let u=lnx [Why?].
Then
du 1
  dx = xdu
dx x
Thus
I 1= 
u2
u3
(lnx)3
xdu =  u 2 du =
c=
c,
x
3
3
I 2= 
1
1
u-1
1
1
xdu =  2 du =
 c =-  c = 
 c,
2
xu
u
-1
u
lnx
I3 = 
u
2
2
xdu =  udu = u3/2  c = (lnx)3/2  c
x
3
3
56
lnx
dx
x
TOPIC 5: CALCULUS

Christos Nikolaidis
CALCULATION BY INSPECTION
If you get used to this simple case of substitution you may directly
write down the result as we are going to explain below. But unless
you feel confident enough follow the whole process of substitution!
Most of the integrals that require substitution have one of the
following forms:
integral
result
 ue dx
= eu + c
 usinudx
=-cosu + c
 ucosudx
= sinu + c
u
u
 u dx
n
 u  u dx
where u is
a function of x.
= lnu + c
=
un+1
+c
n +1
Indeed, the derivatives of the functions in the 2nd column are the
expressions in the 1st column (chain rule). That is why the
substitution method is also known as ANTI-CHAIN rule.
For example,
2x  cosx
 x 2  sinx dx
has the form 
u
dx
u
Hence the result is
lnu + c = ln(x 2 + sinx) + c
Also
(lnx) 2
 x dx
has the form  u  u 2 dx
Hence the result is
u3
(lnx)3
+c=
+c
3
3
57
TOPIC 5: CALCULUS
Christos Nikolaidis
Sometimes a slight modification is needed to have the required
form. For example,
the integral 
x
u
dx is “almost” of the form  dx
x 4
u
2
(since the derivative of x2+4 is 2x and not x).
But we can write
x
1
2x
 x 2  4 dx  2  x 2  4 dx
1
= ln(x 2 + 4) + c
2
[now it has exactly the form 
u
dx ]
u
EXAMPLE 6
 sinxe
Find
cosx
dx ,
Solution
The integral is “almost” of the form  ueudx , since u =-sinx .
Thus
 sinxe
cosx
dx   -sinxecosx dx = ecosx + c
EXAMPLE 7
2
 xcos(3x + 1)dx ,
Find
Solution
The integral is “almost” of the form  ucosudx , since u = 6x . Thus
2
1
1
2
2
 xcos(3x + 1)dx  6  6xcos(3x + 1)dx  6 sin(3x + 1) + c
EXAMPLE 8
I= 
Find
5x 2
dx ,
x3  1
Solution
The integral is “almost” of the form 
Thus
u
dx , since u = 3x 2 .
u
5x 2
5
3x 2
5
dx

dx = ln(x 3 + 1) + c
3
 x3  1

3 x 1
3
58
TOPIC 5: CALCULUS
Christos Nikolaidis
b
5.10 THE DEFINITE INTEGRAL  f(x)dx -AREA BETWEEN CURVES
a
The definite integral is denoted by
b
 f(x)dx
a
where a and b are real numbers within the domain of f.
The calculation is easy:
If
 f(x)dx = F(x) + c ,
then
 f(x)dx = F(x) which means F(b)-F(a)
b
b
a
a
For example,
Definite:
 2xdx  x2 + c
 2xdx =  x  = (32)-(12) = 8
Indefinite:
 (2x  3)dx = x2 + 3x + c
Definite:
 (2x  3)dx = x  3x  = (28)-(0) = 28
Indefinite:
3
2 3
1
1
4
4
2
0
0
Notice: the constant c is omitted in the definite integral! (why?)
EXAMPLE 1
1
 (8x + 12x - 6x + 3)dx = 2x + 4x - 3x + 3x 
3
2
4
3
2
1
-1
-1
= (2 + 4- 3 + 3)- (2- 4- 3- 3)
=8- (-8) = 16
EXAMPLE 2
3
3
3
 x-2 
8
4
4
5
 4
-3
dx
=
8x
dx
=
8
 = - 2  = (- )- (-1) =- + 1 =
2 x 3
2
9
9
9
 -2  2  x  2
3
59
TOPIC 5: CALCULUS
Christos Nikolaidis
 GEOMETRICAL INTERPRETATION
Consider the graph of the straight line f(x)=2x
The area under the line between 0 and 1 (blue triangle) is
but also
1
2 1
0
0
 2xdx =  x  = (1)-(0) = 1.
The area under the line between 0 and 3 (big triangle) is
but also
12
1.
2
3
2 3
0
0
36
9.
2
 2xdx =  x  = (9)-(0) = 9.
The area under the line between 1 and 3 (red region) is 9-1=8.
but also
3
2 3
1
1
 2xdx =  x  = (9)-(1) = 8.
This is not an accident!
In general, if y=f(x) is a “continuous” curve above the x-axis, then
the area between the curve and the x-axis, from x=a to x=b is
equal to the definite integral
b
 f(x)dx
a
We are going to discuss the areas in more detail in the following
paragraph. Now, let us concentrate on the algebraic part of the
definite integral.
60
TOPIC 5: CALCULUS
Christos Nikolaidis
 USE OF GDC
We can use the GDC to find the definite integral.
For Casio CFX we select

MENU

Run-Matrix

MATH [F4]

[F6] and then [F1]
We can also see the result graphically

MENU

Graph (insert the function)

G-Solv [F5]

[F6] and then [F3]

At the moment select  dx [F1]

Select lower and upper limit values (just enter the values)
 PROPERTIES OF THE DEFINITE INTEGRAL
The rules of indefinite integrals are still valid here
b
b
b
a
a
a

 [f(x)  g(x)]dx =  f(x)dx +  g(x)dx

 af(x)dx =a  f(x)dx
b
b
a
a
Moreover,

b
c
c
a
b
a
 f(x)dx +  f(x)dx =  f(x)dx
(use areas to explain why!)
[notice that the limits a,b,c are consecutive]
a
b
b
a

 f(x)dx =   f(x)dx
(accepted by definition)

 f ΄(x)dx = f(x)
(derivative and integral
b
b
a
a
are inverse to each other)

b
b
b
a
a
a
 f(x)dx =  f(y)dy =  f(t)dt = …
[in other words, the variable x is irrelevant!!]
61
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 3
5
Suppose that  f(x)dx  10 . It is also given that f(0)=15, f(5)=3.
0
Then we can also find the following integrals:
5

 2f(x)dx = 20

 f(x)dx = -10

 2 f(x)dx = -5

 (f(x)  4x)dx =  f(x)dx +  4xdx = 10 + 2x  = 10+50 = 60

 (2f(x)  1)dx =  2f(x)dx +  1dx = 20 + x  = 20+5 = 25
0
0
5
0
1
5
5
5
5
0
0
0
5
5
5
0
0
0
2
5
5
0
2
0
2 5
0
5
0

 f(x)dx   f(x)dx =  f(x)dx  10

 f ΄(x)dx = f(x) = f(5)-f(0) = 15- 3 = 12

 f(t)dt =  f(x)dx = 10
5
5
0
0
5
5
0
0
 SUBSTITUTION FOR DEFINITE INTEGRALS
Consider an integral
b
 f(x)dx that requires a substitution u=g(x).
a
The limits a and b refer to x (we say dx).
After substitution dx becomes du, thus the limits a, b are not valid.
In such a case, it would be safe to

find the indefinite integral first (expressed in terms of x)

then evaluate the definite integral.
Let us explain.
62
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 4
I= 
Find
2
0
x
dx ,
x 4
2
Solution
Method A: we find the indefinite integral first
du
du
 2xdx  dx=
, so that
dx
2x
x
x du 1
1
2
 x 2  4 dx =  u 2x = 2 lnu  c = 2 ln(x  4)  c
We use u=x2+4, thus
Therefore,
2
1
1
1 8 1

1
I=  ln(x 2  4) = ln8  ln4 =
ln = ln2
2
2 4 2
2
0 2
Method B: we change the limits
du
du
 2xdx  dx=
dx
2x
Again we use u=x2+4, thus
We also change the limits of the definite integral using u=x2+4
x
u
0
4
2
8
In this case we do not go back to x. Namely,
8
8 x du
x
1
1
1
1

I=  2
=  lnu = ln8  ln4 = ln2
dx = 
0 x
4
u 2x  2
2
2
4
4 2
2
EXAMPLE 5
5
Suppose that  f(x)dx  10 .
0
Then we can also find the following integrals:


8
5
3
0
 f(x - 3)dx =  f(y)dy  10
by using y=x-3,
1 5
f(y)dy  5
2 0
by using y=2x,

2.5
0
f(2x)dx =
63
dy
= 1  dx = dy
dx
dy
dy
= 2  dx =
dx
2
TOPIC 5: CALCULUS
Christos Nikolaidis
 AREA BETWEEN CURVES
b
In fact, the definite integral  f(x)dx is either positive or negative:
a
If y=f(x) is above the x-axis the result is positive.
If y=f(x) is below the x-axis the result is negative
b
b
 f(x)dx = (red area)
 f(x)dx = - (red area)
a
a
Indeed, if you look at the curve y=-2x from x=1 to x=3 then
3
2 3
1
1
 -2xdx = -x  = -9+1 = -8
But of course, we say that the area between the curve and x-axis
from x=1 to x=3 is equal to 8.
64
TOPIC 5: CALCULUS
Christos Nikolaidis
Sometimes we need some extra work to find the limits of the
integral.
EXAMPLE 6
Find the area of the region between the curve y=9-x2 and x-axis
Firstly, we need the roots of the function:
9-x2 =0  x=-3, x=3.
Hence,
3
Area =  (9- x 2 )dx
-3
3

x3 
= 9x- 
3 -3

= (27-9)- (-27 +9) = 18+ 18 = 36
Notice that the curve is symmetric about the y-axis so we can find
the area between x=0 and x=3 and multiply by 2.
3

x3 
Area = 2  (9- x )dx = 2 9x =2(27-9) = 36
0
3 0

3
2
NOTICE for the GDC - graph mode
After G-solv and integral, select the option MIXED - [F4].
Then you may
either enter the limits on yourself
or use the arrows to enter the roots directly
65
TOPIC 5: CALCULUS
Christos Nikolaidis
Suppose that f(x) has both positive and negative values, say
y=f(x)
A
a
C
b
B
c
d
where A,B,C are the areas of the regions shown above.
Then
d
 f(x)dx
a
= A-B+C
Hence,
d
the integral  f(x)dx does not give
a
the total area between the curve and the x-axis.
This is given by the formula
d
AREA =  | f(x) | dx
a
In our example the area is A+B+C. In practice, we have to split this
formula into three integrals
b
 f(x)dx =A
a
c
 f(x)dx =-B
b
d
 f(x)dx =C
c
and then
AREA = A+B+C
NOTICE The GDC - graph mode gives the AREA as well.
After G-solv and integral, select the option MIXED - [F4].
You obtain both,
d
the value of  f(x)dx
a
d
and the AREA  | f(x) | dx
a
66
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 7
Consider the function
f(x) = x 3 - 6x 2 +8x
Find
2
(a)  f(x)dx
0
4
4
(b)  f(x)dx
(c)  f(x)dx
2
0
(d) the total area between the curve and the x-axis within [0,4]
Solution
2
2
(a)  f(x)dx
0
4
4
(b)  f(x)dx
2
4
 x4

=  - 2x 3 + 4x 2  = 0–4 = -4
4
2
4
(c)  f(x)dx
0
 x4

=  - 2x 3 + 4x 2  = 4–0 = 4
4
0
 x4

=  - 2x 3 + 4x 2  = 0–0 = 0
4
0
[it is in fact the sum of the first two integrals above]
(d) The total area is given by
4
A=  x 3 - 6x 2 +8x dx
0
In practice, we find (a) and (b) and add the absolute values
A = 4+4= 8
Notice: your GDC in graph mode finds both
(c) the definite integral = 0
(d) the total area = 4
67
and
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 8
Consider the function
f(x)=sinx,
0  x  2π
Find
π
(a)  f(x)dx
0
(b) 
2π
π
(c) 
f(x)dx
2π
0
f(x)dx
(d) the total area between the curve and the x-axis within [0,2π]
Solution
π
(a)  sinxdx =  cosx ]0π =-cosπ+cos0=1+1=2
0
(b) 
2π
π
(c) 
2π
0
sinxdx =  cosx ]π2π =-cos2π+cosπ=-1-1=-2
sinxdx =  cosx ]02π =-cos2π+cos0=-1+1=0
[it is in fact the sum of the first two integrals above]
(d) The total area is given by
A= 
2π
0
| sinx | dx
In practice, we find (a) and (b) and add the absolute values
A = 2+2= 4
Notice: your GDC in graph mode, estimates directly both
(c) the definite integral = 0
and
68
(d) the total area = 4
TOPIC 5: CALCULUS

Christos Nikolaidis
AREA BETWEEN TWO CURVES
Consider the graphs of two functions y=f(x) and y=g(x), such that
f(x)  g(x) for any x
y=f(x)
y=g(x)
a
b
The area between the two curves from x=a to x=b is given by
b
 [f(x)- g(x)]dx
a
b
while
 f(x)dx gives the area between y=f(x) and x-axis
 g(x)dx gives the area between y=g(x) and x-axis
hence
their difference gives the shaded area requested!
Indeed,
a
b
a
EXAMPLE 9
Find the shaded area below:
6
Area =  [(
2
6
1
1
x  4) - x]dx =  (4 - x)dx
2
2
2
6

x2 
= 4x 
 = (24-9)-(8-1) = 15-7 = 8
4 2

69
TOPIC 5: CALCULUS
Christos Nikolaidis
In general, the area between two curves is given by
b
 | f(x)- g(x) | dx
a
Consider the following situation
 METHODOLOGY

Find the intersection points by solving the equation f(x)=g(x)
(in our example: x=a, x=b, x=c)

Split the integral appropriately: for each interval, determine
which curve is “above” and which curve is “below”
b
c
a
b
(in the example:  [f(x)- g(x)]dx +  [g(x)- f(x)]dx )
EXAMPLE 10
Find the area enclosed by the graphs f(x)=x2 and g(x)=x+2

Intersection points:
f(x)=g(x)

x2=x+2

x2-x-2=0
70
TOPIC 5: CALCULUS

Christos Nikolaidis
The shaded area is
2
 x2
x3 
[(x

2)x
]dx

(x

2x
)dx
=


2x


-1
-1
3 -1
2
8
1
1
10 7
9
= (2+4- )- ( -2+ ) =
+
=
3
2
3
3
6
2
2
2
2
2
Mind carefully the region required.
EXAMPLE 11
Consider the curve
y x
We define two regions
A: among the curve y  x , x-axis and the line x=9.
B: among the curve y  x , y-axis and the line y=3.
The corresponding areas are given by
9
9
0
0
A   x dx   x
9


9

B   3  x dx   3- x
0
0
1/2
1/2
9
2
2

dx   x 3/2   93/2  0  18
3
0 3
9
2
2




dx  3x  x 3/2    27  93/2   0  9
3
3

0 


or
B  ( Area of rectangle)  A  27 - 18  9
71
TOPIC 5: CALCULUS
Christos Nikolaidis
5.11 KINEMATICS (DISPLACEMENT-VELOCITY-ACCELERATION)
Consider a body moving along a straight line. The body is free to
move forwards and backwards. The displacement s is the distance
|OA| from a fixed point O (the origin).
s
O
A
The displacement s is given as a function of time. For example
s=t2-4t+3
means that
at time t=0 the displacement is 3 units from the fixed point O
at time t=1 the displacement is 0, (the body goes back to point O)
at time t=2 the displacement is -1, (the body is before point O)
at time t=3 the displacement is 0, (at point O again)
at time t=4 the displacement is 3, (it is moving forward)
Then
Velocity
= rate of change of displacement:
Acceleration = rate of change of velocity:
Notice also that a is the second derivative of s.
Displacement
Velocity
Acceleration
s
v
a
derivative
72
v=
ds
dt
a=
dv
dt
a=
d2 s
dt 2
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 1
Consider
s= t3-12t+15
representing the motion of a particle along a straight line, where
the displacement s is given in m (meters),
the time t is given in sec (seconds).
Then
ds
dv
=3t2-12
a=
=6t
dt
dt
Notice that v is measured in m/sec while a is measured in m/sec2.
v=
For example,
at time t=1,
s=4m
v=-9m/sec
a=6m/sec2
at time t=3,
s=6m
v=15m/sec
a=18m/sec2
Notice also,
The body is stationary when the velocity is 0
Hence, let us solve
v=0

3t2-12=0

t2=4

t=2sec
(notice that time is always positive, thus we reject t=-2)
At that time (t=2),
s=-1m
v=0m/sec
a=12m/sec2
73
TOPIC 5: CALCULUS
Christos Nikolaidis
NOTICE
If s>0, the body is to the right of the fixed point O.
If s<0, the body is to the left of the fixed point 0.
If s=0, the body is at the fixed point 0.
If v>0, the body is moving to the right
If v<0, the body is moving to the left
If v=0, the body is stationary (it changes direction)
If a>0, the body accelerates
If a<0, the body decelerates
If a=0, the velocity is stationary (probably maximum speed)
The motion may be given as a graph of s with respect to t
The displacement s from the fixed point O is the y-coordinate for
each t-value (time); It can be
positive (after the fixed point O);
negative (before the fixed point O); or
zero (at the fixed point O).
74
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 GOING BACKWARDS (BY USING INTEGRATION)
If we are given the acceleration of a moving body we can find the
velocity and the displacement by using integration.
Displacement
Velocity
Acceleration
s
v
a
derivative
integral
However, in this opposite direction we must be given some extra
information in order to estimate any emerging constant c.
EXAMPLE 2
Let v=12t2-2t. Find the acceleration and the displacement, given
that the initial displacement is 5m.
Solution
a=
dv
=24t-2 ms-1
dt
s=  vdt =  (12t 2 - 2t)dt =4t3-t2+c
but s=5, when t=0, thus c=5. Hence, s=4t3-t2+5 m
EXAMPLE 3
Let a=12t. Find the displacement, given that the moving body
starts from rest; the initial displacement is 5m.
Solution
v=  adt =  12tdt =6t2+c
but when t=0, v=0 (at rest!), thus c=0. Hence, v=6t2
s=  vdt =  6t 2 dt =2t3 +c
but when t=0, s=5, thus c=5. Hence, s=2t3 +5 m
Mind the difference between the displacement and the distance
travelled of a moving body.
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 DISPLACEMENT vs DISTANCE TRAVELLED
Suppose that the velocity v is given in terms of t. Then
Displacement from O
s=
Displacement from t1 to t2
S =
 vdt
t2
 vdt
t1
Distance travelled from t1 to t2
d =
t2
 v dt
t1
EXAMPLE 4
The velocity of a moving body is given in ms-1 by v = 12 – 3t 2
The initial displacement from a fixed point O is 1m.
Then

The displacement of the body from O is given by
s=  vdt =12t-t3+c
Since s=1 when t=0, we obtain c=1 and so s=12t-t3+1
Thus, after 2 seconds, s = 17m. After 3 seconds s = 10m

The displacement in the first 2 seconds is
2
2
0
0
2
S=  vdt   (12- 3t 2  1)dt  12t  t 3  t  =16m
0
while the displacement in the first 3 seconds is
3
3
0
0
3
S=  vdt   (12- 3t 2 )dt  12t  t 3  =9m

0
The distance travelled in the first 3 seconds is
2
2
2
0
0
0


2
d=  v dt   12 - 3t 2 dt   (12 - 3t 2 )dt  12t  t 3 0 =16m
while the distance travelled in the first 3 seconds is
3
3
2
3
0
0
0
2
d=  v dt   12 - 3t 2 dt   (12- 3t 2 )dt -  (12- 3t2 )dt =16+7=23m
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Explanation
O
A(t=0)
0m 1m
C(t=3)
B(t=2)
10m
17m
Time
Displacement
Displacement
Distance
Velocity
t
from O
from t=0 to t
travelled
v
s = 1m
S = 0m
d=0m
v=12
s = 17m
S = 16m
d=16m
v=0
s = 10m
S = 9m
d=23m
v=-15
Initially t=0
(Position A)
after t=2 sec
(Position B)
after t=3 sec
(Position C)
In other words, the moving body
starts from A
is moving forward in the first two seconds (to position B)
changes direction at B (v=0)
goes back to position C.
If we draw the graph of v against t,
v
t
1
2
3
The displacement S from t=0 to t=3 is the definite integral from
t=0 to t=3 (area above the t-axis minus the area below the t-axis).
The distance travelled from t=0 to t=3 is given by the total area
from t=0 to t=3 between the curve and t-axis (shaded area above)
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TOPIC 5: CALCULUS
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EXAMPLE 5
Let
v  4t  t 2
Given that the initial displacement is 10m find the displacement
and the distance travelled
a) during the first after 3 sec
b) during the first after 6 sec.
Solution
Let us solve first the equation v=0
v  0  4t  t 2  0  t  0 or t  4
So the body changes direction after 4 sec.
(a) During the first 3 seconds the displacement and the distance
travelled coincide.
3
3

t3 
S = d   (4t- t )dt =  2t 2    9
3 0

0
3
(b) During the first 6 seconds the body changes direction.
The displacement is
6
6

t3 
S   (4t- t )dt =  2t 2    0
3 0

0
3
so it goes back to the initial position.
For the distance travelled, we have to split the integral:
4
4
 2 t3 
32
=
(4t
t
)dt

 2t   
0
3
3

0
2
6
6
 2 t3 
32
=
(4t

t
)dt
 2t    
4
3 4
3

2
Hence,
d=
32
32 64
+
= 21.3m

3
3
3
Notice The GDC directly gives the results. For example, for (b)
6
6
S   (4t- t 3 )dt  0 m
d   4t  t 2 dt  21.3 m
0
0
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TOPIC 5: CALCULUS
Christos Nikolaidis
ONLY FOR
HL
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Christos Nikolaidis
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Christos Nikolaidis
5.12 CONTINUITY AND DIFFERENTIABILITY
 CONTINUITY
In paragraph 5.1 we have seen that
lim f(x)  7 = f(2)
for f(x) = 2x+3,
for f(x) 
x 2
sinx
x
lim f(x)  1 , but f(0) is not defined.
x 0
If we observe the graph of f(x) 
sinx
x
y
1
x
-5
5
we notice a “discontinuity” at x=0.
It’s worth it to see another similar situation.
f(x) 
x2  4
x2
The function is not defined at x=2. However x=2 is not a vertical
asymptote. It is interesting to see what happens to f(x) as x 2.
x
f(x)
1.9
3.9
1.99
3.99
1.999
3.999
2.001
4.001
2.002
4.002
It seems that lim f(x)  4 .
x 2
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Indeed, look at the graph of f (there is an “empty” point on it)
y
6
5
4
3
2
1
x
-3
-2
-1
1
2
3
4
-1
When x approaches 2, the value f(x) approaches 4. Thus
f(2) is not defined
lim f(x)  4 .
but
x 2
In fact, we can simplify the function as
f(x) 
x 2  4 (x  2)(x  2)

 x  2,
x2
x2
where x  2
That is why we obtain the graph of the straight line y=x+2 with
some “discontinuity” at x=2. Moreover,
x2  4
 lim (x  2)  4
x 2 x  2
x 2
lim
We say that a function is continuous at x=a, when

The value f(a) exists;

The limit lim f(x) exists;

lim f(x)  f(a)
x a
x a
For the continuity at any particular point we must check all three
presuppositions.
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Most of the known functions are continuous everywhere (since they
look like uninterrupted curves!). For example, lines, quadratics,
polynomials in general, exponentials are continuous functions.
Remember that at some x=a we may have different side limits
lim f(x) and lim f(x)
x a
x a
If they are equal, say lim f(x) = lim f(x) = b, then we can say that
x a
x a
limf(x) = b
xa
It is worthwhile to see the following examples of “step” functions to
further clarify the notions of limit and continuity.
EXAMPLE 1
2x  1,
f(x)  
7,
Let
if x  2
if x  2
y
7
6
5
4
3
2
1
x
-3
-2
-1
1
2
3
4
5
-1

limf(x)  5

f(2) =7.

But lim f(x)  f(2)
x2
[when x approaches 2,the value f(x) approaches 5];
x 2
Thus the function is not continuous at x=2.
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EXAMPLE 2
Consider the function
y
7
x 2 , if
f(x)  
 5, if
x2
6
x2
5
4
3
2
1
-4
-3
-2
x
-1
1
2
3
4
5
6
7
6
7
-1
We can see that f(2)=4 but lim f(x) does not exist.
x 2
[In fact, only side limits exist: lim f(x)  4 and lim f(x)  5 ]
x 2
x 2
Therefore, the function is not continuous at x=2.
(thus the functions is not continuous in general).
EXAMPLE 3
Consider the function
y
7
x 2 , if x  2
f(x)  
 4, if x  2
6
5
4
3
2
1
-4
-3
-2
-1
x
1
2
3
-1
We can see that f(2)=4 and also lim f(x)  4 .
x 2
Since lim f(x)  f(2) the function is continuous at x=2.
x 2
(in fact the function is continuous everywhere).
84
4
5
TOPIC 5: CALCULUS
Christos Nikolaidis
 THE FORMAL DEFINITION OF THE DERIVATIVE
Let y=f(x) be a continuous curve and A(x,f(x)) some point on it:
y
B
f (x+h)
f (x)
A
x
x
x+h
We select a neighboring point B with
x-coordinate = x+h
(where h is very small)
y-coordinate =f(x+h)
As we move from point A to point B, the rate of change is
Δy f(x  h)- f(x)

Δx
h
If we let h become very small, that is h0, the result will be the
rate of change at point A, that is the derivative f (x) .
f(x  h)- f(x)
h 0
h
f (x) = lim
Let us apply this formula to the function f(x) = x2.
f(x  h)- f(x)
h0
h
f (x)  lim
(x  h) 2  x 2
h0
h
 lim
x 2  2xh  h 2  x 2
h0
h
 lim
 lim
h0
h(2x  h)
h
 lim (2x  h)  2x
h 0
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TOPIC 5: CALCULUS
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Therefore,
f (x)  2x
If we apply the definition for f(x)=xn we will find that
f(x  h)- f(x)
 nx n-1
h 0
h
f (x)  lim
(exercise!)
EXAMPLE 4
Show from first principles (that is by using the formal definition),
that the derivative of the function
f(x)  x 3  2x
is
f (x)  3x 2  2 .
Solution
f (x)
f(x  h)- f(x)
h0
h
 lim
[(x  h)3  2(x  h)]  [ x 3  2x]
h0
h
 lim
x 3  3x 2 h  3xh 2  h 3  2x  2h  x 3  2x
h0
h
 lim
3x 2 h  3xh 2  h 3  2h
h0
h
 lim
h(3x 2  3xh  h 2  2)
h0
h
 lim
 lim (3x 2  3xh  h 2  2)
h0
Now, we are able to set h=0 and obtain,
f (x)  3x 2  2
If f (x) exists we say that f(x) is differentiable at x. We also say that
f(x) differentiable if it f (x) exists for any point x of the domain.
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TOPIC 5: CALCULUS
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Notice
f differentiable at x

f continuous at x
The opposite is not necessarily true: that is, the function may be
continuous at some point x but not differentiable at this point.
Indeed, let us see again the step function of example 3
y
7
x 2 , if x  2
f(x)  
 4, if x  2
6
5
4
3
2
1
-4
-3
-2
-1
x
1
2
3
4
5
6
7
-1
The function is continuous at x=2 but it is not differentiable at x=2.
Informally speaking, this is because
There is a “corner” point: we cannot draw a tangent line at x=2
or otherwise,
The curve is continuous at x=2 but is not “smooth” at this point
But let us give a more formal proof.
Method 1 (without mentioning the definition from first principles)
At x=2 we find the “side” derivatives of f:
f- (x)  (x 2 )  2x , thus
f- (2)  4
f (x)  (4)  0 ,
f  (2)  0
thus
Since f- (2) and f (2) differ, f (2) does not exist.
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TOPIC 5: CALCULUS
Christos Nikolaidis
Method 2 (from first principles)
At x=2 we find the “side” derivatives of f:
f(2  h)- f(2)
(2  h) 2 - 4
4  4h  h 2 - 4
 lim limh 0
h0
h 0
h
h
h
f- (2)  lim-
4h  h 2
 lim- (4  h)  4
h0
h 0
h
 limwhile
f(2  h) - f(2)
4- 4
 lim
 lim (0)  0
h 0
h0
h0
h
h
f  (2)  lim
Since f- (2) and f (2) differ, f (2) does not exist.
Let us slightly modify this function:
EXAMPLE 5
 x2,
if x  2
f(x)  
4x - 4, if x  2
We firstly show that f is continuous at x=2:



lim f(x)  2 2  4
x 2
lim f(x)  8  4  4
x 2 
f(2)=4
Therefore lim f(x)  4  f(2) and the functions is continuous at x=2.
x 2
We next show that f is differentiable at x=2:

f- (x)  (x 2 )  2x ,
thus
f- (2)  4

f (x)  (4x - 4)  4 ,
thus
f  (2)  4
Therefore f (2)  4 and the functions is differentiable at x=2.
Notice: let’s also show that f  (2)  4 from first principles:
f(2  h)- f(2)
[4(2  h)- 4]- 4
4h
 lim
 lim (
)  lim (4)  4
h 0
h0
h0
h0
h
h
h
f  (2)  lim
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TOPIC 5: CALCULUS
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We will see the same example in a different version:
EXAMPLE 5
 x2,

f(x)   a,
 bx  c,

if x  2
if x  2
if x  2
Find a, b and c given that the function is (continuous and)
differentiable.
Solution
The function is continuous and differentiable for any x≠2.
We check continuity and differentiability at x=2.
Continuity:



lim f(x)  2 2  4
x 2
lim f(x)  2b  c
x 2 
f(2)=a
Since f is continuous at x=2:
4  2b  c  a
(1)
Differentiability:

f- (x)  (x 2 )  2x ,
thus
f- (2)  4

f (x)  (bx  c)  b ,
thus
f (2)  b
Since f is differentiable at x=2:
b4
(1) and (2) give
a=4, b=4, c=-4
89
(2)
TOPIC 5: CALCULUS
Christos Nikolaidis
5.13 L’HÔPITAL’s RULE (for HL)
 A FIRST DISCUSSION
We know that
0
0
5
5
0
is not defined
However, when x tends to 0,
5
approaches either + or -
x
But what about
0
?
0
Consider a function of the form
f(x)
g(x)
if
if
f(x)→0
g(x)→5
f(x)→5
g(x)→0
then
then
f(x)
tends to 0
g(x)
x- 5 0
 0
x 5
x
5
e.g
f(x)
tends to + or –
g(x)
lim
5
 +
x 5 x - 5
lim
e.g
lim
5
x 5  x - 5
 –
But again, what happens when both f(x) and g(x) tend to 0?
The result could be anything: 0 or + or – or any real number!
For example, when x→0, all the functions below have the form
however
x3
 0,
x 0 x
lim
lim
x
x 0 x 3
 ,
lim
x 0
sinx
 1,
x
lim
x 0
2sinx
2
x
(check these functions on your GDC near x=0).
That is why we say that
0
is an indeterminate form.
0
Another indeterminate form is

.

90
etc
0
,
0
TOPIC 5: CALCULUS
Christos Nikolaidis
f(x)

0
IS NOT OF THE FORM
or
g(x)
0

 CASES WHERE
Here we deal with limits of the form
f(x)
,
x a g(x)
f(x)
,
x   g(x)
lim
2x  3 5
lim

x 1 3x  5
8
lim
x  
2
0
3x  5
lim

0
or
we can easily find the
0

If the fraction is not of the form
result. For example,
f(x)
x -  g(x)
lim
1
x 5
lim
x  
1 7
7
x
x
e
lim
0
x -  x  3
5
x 3
0
 0
2
x 3 x
3 9
lim
[
2
0
and
]


lim
2x  7
 
x  
3
lim
2x  7
 
x  
ex
[


and  ]
3
0
lim
2
 
x 3 (x  3) 2
lim
2
 
x 3 (x  3) 2
[
2
-2
and  ]

0
0
2
 
x 3 x  3
lim
2
 
x 3 x  3
[
2
2
and  ]

0
0
lim
 THE INDETERMINATE FORMS
0

or
0

In some cases the answer is easy. For example

:

4x  7
 2,
x   2x  3
lim
4x  7
 0,
x   2x 2  3
lim
4x 2  7
 
x   2x  3
lim
[remember the discussion about horizontal asymptotes]
0
:
0
x(x - 3)
x 2 - 3x
 lim
 lim x  3
x 3
x

3
x 3
x- 3
x- 3
lim
It is also known that
sinx
1
x 0
x
lim
(it is
0
)
0
For more complicated cases the following theorem helps; we simply
need the derivatives of f(x) and g(x).
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TOPIC 5: CALCULUS
Christos Nikolaidis
L’ Hôpital’s rule:
If
f(x)→0
or
g(x)→0
provided that lim
f(x)→±∞
lim
g(x)→±∞
f(x)
f  (x)
 lim
g(x)
g  (x)
f  (x)
exists.
g  (x)
EXAMPLE 1

 
(4x  7)
4x  7   
4
lim
 lim
 lim  2
x   2x  3
x   (2x  3) 
x   2
0
 
0

 

Remember to indicate the form above = :  or  .
Sometimes we need to apply L’Hôpital’s rule more than once.
EXAMPLE 2

 

 
2x 2  4x  7   
4x  4   
4 2
lim
lim

 lim

2
x   5x
x   10x  3
x   10
5
 3x  2
EXAMPLE 3
0
 
0
 
ex  x  1  0 
ex  1  0 
ex 1
lim
lim
lim



x 0
x0
x   2
2x
2
x2
EXAMPLE 4
Find the horizontal asymptotes of f(x) 

when x  
3e 2x  2
e 2x  1

 
3e 2x  2   
6e 2x
6
lim 2x
 lim
 lim  3
2x
x   e
x   2e
x   2
1

when x - , the limit is not of the form
3e 2x  2  2

 2
x   e 2x  1
1
H.A. y=3

since e 2x  0 .

H.A. y=-2
lim
92
TOPIC 5: CALCULUS
Christos Nikolaidis
 OTHER INDETERMINATE FORMS
The following are also indeterminate forms
0 ,
 ,
1
For example, look at the following 0   forms:
 1
lim  x    lim 1  1
x 0 
x  x 0
1

lim  x 2    lim x  0
x 0 
x  x 0
1 
1

lim  x  3   lim 2  
x 0 
x  x 0 x
More complicated forms can be transformed to the form
and thus answered by using L’Hôpital’s rule.
0

or
0

EXAMPLE 5
When x→0 then lnx→-∞. Thus lim (xlnx ) is of the form 0   .
x 0
But
1

 
lnx   
lim (xlnx )  lim
 lim x  lim (-x)  0
x 0
x 0 1
x 0
1 x 0
 2
x
x
EXAMPLE 6
The limit lim
x  
lim
x  
 x  1  2x  is of the form    . But
 x  1  2x 
 x  1  2x   lim  x  1  x 2x
 1  2x
x  
 lim
x  

 

 lim
x  
x  1- 2x
1- x
 lim
x


x  1  2x
x  1  2x
-1
 
1
1

2 x 1
2x
93
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 7
We will show that
x
1

lim  1    e .
x   
x
It has the indeterminate form 1  . But

x
1
x

1
xln  1 
ln  1  
1

x
e  x
1    e 
x



Let’s find the limit of the exponent xln 1 
1
1
 ; it has the form 0   :
x
 1 


1 0
1   x2 


ln 1    0 
1  
1
1
x  
x




lim

lim
 lim
1
lim xln 1  
x


x


x


x  
1
1
1
x


 2
1  
x
x
x

Hence

x
1
xln  1 
1

lim  1   = lim e  x  =e1=e
x   
x  x 
NOTICE

x   

In a similar way we can show that lim  1 

As we said earlier we accept the limit lim
x
a
a
 e .
x
sinx
 1 as known.
x 0
x
Although it is of the form
sinx
lim
x 0
x
0
 
0
0
and the rule would give
0
cosx
(sinx)
 lim
1
x 0
x 0
x
1
 lim
we cannot use L’Hôpital ! This is because the proof of the fact
sinx   cosx
already uses the limit lim
x 0
(by first principles)
sinx
1.
x
The proof of this limit in particular uses other trigonometric
techniques and it is beyond the scope of this course.
94
TOPIC 5: CALCULUS
Christos Nikolaidis
5.14 IMPLICIT DIFFERENTIATION – MORE KINEMATICS (for HL)
We are very familiar with functions of the form y=f(x). However, in
many real applications, we are not given a clear function where y
is expressed in terms of x, but a more complicated relation which
involves x and y.
Consider for example the relation
x2+y2=1
Actually, this relation represents a circle of radius 1 on the
Cartesian plane:
y
x
-1
1
(i.e. the pairs (x,y) that satisfy this relation form that circle)
Obviously this is not the graph of a function (as a vertical line may
cross the graph at two points). However, if we solve for y we obtain
y2=1-x2  y=  1  x 2
that is, two different functions together:
y= 1  x 2
is the semicircle above the x-axis
y=  1  x 2
is the semicircle under the x-axis
The question is
What is the derivative y΄=
95
dy
dx
?
TOPIC 5: CALCULUS
Christos Nikolaidis
Case 1: If y= 1  x 2 the derivative is y΄ =
Case 2: If y=  1  x 2
 2x
2
=-
x
2 1x
1  x2
 2x
x
the derivative is y΄ = 
=
2 1  x2
1  x2
We can observe that in both cases the result is equal to y΄=-
x
y
A more elegant way to obtain the same result is the following:
Consider again
x2+y2=1
Differentiate both sides with respect to x. Do have in mind though
that y is a function of x, so the derivative of y2 is not 2y but 2yy΄
(chain rule). Hence, the differentiation gives
2x+2yy΄= 0  2yy΄=-2x  y΄=-
x
y
Notice that it is not necessary to solve the initial equation for y
(sometimes this may be very difficult or even impossible!). We just
differentiate both sides with respect to x and then solve for y΄. This
process is known as implicit differentiation.
EXAMPLE 1
Find y΄=
dy
dx
2x 2  x 2 y3  y 2
if
(notice that solving for y seems to be a nightmare!!!)
Solution
The implicit differentiation gives
4x+2xy3+x23y2y΄=2yy΄
Now we simply solve for y΄:
4x+2xy3=2y y΄-3x2y2y΄

4x+2xy3=(2y-3x2y2)y΄

y΄=
4x  2xy3
2y  3x 2 y 2
96
TOPIC 5: CALCULUS
Christos Nikolaidis
Ok, you may complain that the result is not a clear expression of x
as usual but involves x and y as well!!! However this is not a
problem in general!
EXAMPLE 2
Consider again the equation of the circle
x2+y2=1
Find the tangent lines
1 3
,
)
2 2
(confirm that this point lies on the circle)
a) at the point (x,y)= (
b) at the points of the circle with x=0
Solution
First, we need the derivative y΄=
dy
dx
We have seen that
y΄=a) At (x,y)= (
1 3
,
) , it is
2 2
:
x
y
m=y΄=-
x
1/2
1
3
===y
3
3/2
3
Hence, the tangent line has the form
y =But
3
x+c
3
3 1
3
3
3
4 3
+c=
+
 c=
 c=
3 2
2
2
6
3
So that the tangent line is
-
3
4 3
x+
3
3
b) For x=0 we obtain two values for y:
y =
x2+y2=1  y2=1  y=  1

At (x,y)=(0,1), the gradient is m=y΄=0 and the tangent line is
y= 0x+c, or finally y=1

At (x,y)=(0,-1), the gradient is m=y΄=0 and the tangent line is
y= 0x+c, or finally y=-1
97
TOPIC 5: CALCULUS
Christos Nikolaidis
[Indeed, notice that A(0,1) and B(0,-1) are the highest and the
lowest points of the circle respectively
y
y=1
A
x
-1
1
y=-1
B
Apparently, the tangent lines at those points are the horizontal
lines y=1 and y=-1]
EXAMPLE 3
Let
x2+y+xcosy=π.
Find the tangent and the normal lines at x=0.
Solution
Firstly, for x=0 we obtain y=π, so the given point is (0,π).
Implicit differentiation gives
2x  y΄  cosy  x( siny)y΄  0
 y΄  (xsiny)y΄  -2x  cosy
- 2x  cosy
 y΄ 
1  xsiny
At (x,y)=(0,π), the gradient is
m=y΄=1.
The tangent line has the form y=x+c. By using the point (0,π) we
find c=π and finally
y=x+π
The normal line has the form y=-x+c. By using the point (0,π) we
find c=π and finally
y=-x+π
98
TOPIC 5: CALCULUS
Christos Nikolaidis
 MORE ON KINEMATICS
Remember the following scheme
Displacement
Velocity
Acceleration
s
v
a
derivative
integral
In this situation, s,v,a are all given in terms of time t.
For example, given that
v=6t2
then
a =
dv
=12t
dt
However, the velocity (v) is sometimes given as a function of s,
instead of t. For example, v=3s2
The following formula for acceleration may be derived (using the
chain rule)
a
dv dv ds dv


v
dt ds dt ds
Concentrate on
a v
dv
ds
EXAMPLE 4
Let
v=3s2
Then
a= v
dv
= (3s2)(6s)=18s3
ds
In fact, we use implicit differentiation on v=3s2 with respect to t:
a=
dv
ds
= (6s)
= (6s) v=(6s) (3s2)=18s3
dt
dt
99
TOPIC 5: CALCULUS
Christos Nikolaidis
5.15 RATE OF CHANGE PROBLEMS (for HL)
In many applications we have quantities depending on time t. If A
is a function of t we know that
dA
= rate of change of A
dt
The rate of change of y=A2 is given by
dy dy dA
dA

 2A
dt dA dt
dt
The rate of change of y=A3 is given by
dy
dA
 3A 2
dt
dt
In general, the rate of change of any function of A, say y=f(A) is
given by
dy
dA
= f (A)
dt
dt
a) We are given a relation between two quantities A and B. Any
change according to time in one of them implies a change in the
dA
dB
and
, the rates
dt
dt
of change of A and B, can be found by implicit differentiation
other one as well. The relation between
with respect to time t:
We differentiate both parts of the relation;
dA
;
dt
dB
whenever we differentiate B we multiply by
.
dt
whenever we differentiate A we multiply by
For example
If
A=2B3
then
If
A=2B+lnB
then
If
A2=2B3
then
If
sinA=
3
B
then
100
dA
dB
 6B 2
dt
dt
dA
dB 1 dB
2

dt
dt B dt
dA
dB
2A
 6B 2
dt
dt
cosA
dA
3 dB
 2
dt
B dt
TOPIC 5: CALCULUS
Christos Nikolaidis
b) We are given a relation between three quantities A,B and C. The
dA dB
dC
,
and
, can be
dt dt
dt
found by implicit differentiation with respect to time t.
relation between the rates of change
For example
If
A=2B3+4C2
then
dA
dB
dC
 6B 2
 8C
dt
dt
dt
If
A=B2C3
then
dA
dB
dC
 2BC 3
 3B 2 C 2
dt
dt
dt
If
A3+B2=5
then
3A2
If
cosA=B+5BC
then
-sinA
dA
dB
+ 2B
0
dt
dt
dA dB
dB
dC

 5C
 5B
dt dt
dt
dt
In problems involving rates of change we work as follows
Methodology
The problem usually refers to the rates of change of two quantities.
One rate is given, one rate is required (usually at some instant).
1. Determine the two quantities A and B
dA
dB
is given and
is required)
dt
dt
2. Find the general relation between A and B
(Say that
(*)
dA
dB
and
(**)
dt
dt
4. If the question mentions a specific instant (say for a
3. Find the relation between
particular value of B)

We use (*) to find the value of A (if necessary).

We substitute all known values in (**) to find
dB
dt
Notice: Be careful, the values of A or B at some particular instant
are used only in the final step 4 of substitution
Work similarly if three or more quantities are involved.
101
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 1
Consider a square object which is expanding. If the side of the
object increases in a constant rate of 2ms-1 find the rate of change
of its area, at the instant when the side is 10m.
Solution
x= side,
dx
=2m/sec
dt
A= area,
dA
=?
dt
x
The relation between A and x is:
A=x2
Hence,
Therefore, when x=10m
dA
dx
= 2x
dt
dt
dA
=2×10×2=40m2/sec
dt
EXAMPLE 2
Consider an expanding sphere. If the volume increases in rate
5cm3/sec find the rate of change of its radius r,
(i) when r = 3 cm
(ii) when the volume reaches 36π cm3
Solution
dV
dr
=5cm3/sec,
=?.
dt
dt
The relation between V and r is given by V=
(i) when r = 3,
dV
dr
=4πr2
dt
dt
4
πr 3 . Hence,
3
dr
dr
5
=
m/sec.

dt
dt 36π
4
(ii) when V=36π, the original relation gives 36π= πr 3  r = 3cm.
3
Therefore, the answer is the same as above.
5=4π32
102
TOPIC 5: CALCULUS
Christos Nikolaidis
Have in mind that speed is also a rate of change. When a moving
body A has speed 5m/sec, that means that the distance x from
some fixed point O changes in rate 5m/sec
O
x
A
dx
 5m/sec
dt
dx
If A is moving to the left, x decreases so
 5m/sec
dt
If A is moving to the right, x increases so
EXAMPLE 3
Two cars, A and B, are traveling at 60 km/h and 70 km/h
respectively, on straight roads, as shown in the diagram below.
A
x
O
y
z
B
At a given instant both cars are 100 km away from O. Find, at
that instant, the rates of change
(a) of the distance between the cars.
ˆΒ
(b) of the angle θ  ΟΑ
Solution
dx
dt
dy
 60km/h
dt
dz
 70km/h
dt
dθ
?
dt
When x  y  100 , then z  100 2 and θ 
z 2  x 2  y2
(a) Relation between x, y, z:
Hence
2z
dz
dt
 100 2
 2x
dz
dt
dx
dt
 2y
dy
dt
z
dz
dt
π
4
x
dx
dt
 100(60)  100(70) 
103
?
y
dz
dt
dy
dt
 5 2km/h
TOPIC 5: CALCULUS
Christos Nikolaidis
(b) Relation between x, y, θ
dy
Hence
dt

dx
dt
tanθ  xsec 2 θ
 -70  60tan

tanθ 
:
y
 y  xtanθ
x
dθ
dt
π
π dθ
 100sec 2
4
4 dt
dθ
 0.65 rad/h
dt
Let us see a final example with three variables.
EXAMPLE 4
It is given that
1 2
r h  2r 3
3
Find the rate of change of h when r=3 and h=6, under two
A
circumstances:
a) h is always double of r and
dA
dh
 30 and
8
dt
dt
Solution
dA
 30
dt
b)
a) Since h=2r, the original relation becomes A 
Hence
2 3
8
r  2r 3  r 3
3
3
dA
dr
 8r
dt
dt
Therefore, when r=3
dr
dr 5


dt
dt 4
b) By implicit differentiation on the original relation we obtain
30  24
dA 2
dr 1 2 dh
dr
 rh
 r
 6r 2
dt 3
dt 3
dt
dt
Therefore, when r=3 and h=6,
30  12
dr
dr
dr
dr
1
 24  54
 6  66


dt
dt
dt
dt 11
104
TOPIC 5: CALCULUS
Christos Nikolaidis
5.16 FURTHER INTEGRATION BY SUBSTITUTION (for HL)
We have seen the simple case of substitution u=f(x), when the
derivative of f(x) is part of the integrand.
For example,
I= 
f  (x)
dx = ln(f(x) + c
f(x)
Indeed,
Let u = f(x)
Then 
du
du
 f (x)  dx 
dx
f (x)
and
I= 
f  (x) du
1
  du  lnu  c = ln(f(x) + c
u f  (x)
u
In this case we can omit the whole process and give directly the
result.
EXAMPLE 1
Find
A= 
cosx
dx ,
sinx  1
B= 
x
dx
x 1
2
Solution
Since the derivative of u=sinx+1 is cosx
A= 
cosx
dx  ln(sinx  1)
sinx  1
For B, the derivative of u=x2+1 is 2x.
We slightly modify the integral to obtain the form 
B= 
x
1
2x
1
dx   2
dx  ln ( x 2  1)  c
2 x 1
2
x 1
2
105
f  (x)
dx
f(x)
TOPIC 5: CALCULUS
Christos Nikolaidis
Similarly,
e
f(x)
 f  (x)dx = ef(x) + c
 sinf(x)  f  (x) dx =-cosf(x) + c
 cosf(x)  f  (x) dx =sinf(x) + c
n
 f(x)  f  (x) dx =
f(x)n 1
c
n 1
etc.
EXAMPLE 2
 sinx e
2
x e
cosx
x 3 5
dx = e cosx  c
dx =
3
1
1 3
3x 2 e x 5 dx = e x  5  c

3
3
1
2
2
1
2
 xsin(7x  3)dx = 14  14xsin(7x  3)dx =  14 cos(7x  3)  c
2
5
 5x(x  3) dx =
2
 3x x  3dx =
5
5 (x 2  3)6
5
2
5
=
2x(x

3)
dx
c=
(x 2  3)6  c

2
2
6
12
1
3
32 2
2x(x 2  3) 2 dx =
(x  3)3/2  c = (x 2  3)3/2  c

2
23
(lnx) 3
(lnx)4
1 1
1 (lnx) 4
3
dx
=
(lnx)
dx
=
=

c
c
 2x
2x
2 4
8
Remark.
If you don’t feel confident to find directly the result, you may
always follow the detailed procedure of substitution.
106
TOPIC 5: CALCULUS
Christos Nikolaidis
We are going to see two more cases of substitution.
CASE 1: We reuse u=f(x) to get rid of all remaining x’s.
Consider
I=  x 3 x 2  3dx
Let’s use the substitution
u=x2+3
(despite the fact that x3 is not the derivative of u!)
Then
du
du
=2x  dx=
dx
2x
Thus,
I=  x 3 u
du 1
=  x 2 udu
2x 2
Well, the result still contains x2, but
u=x2+3x2=u-3
and thus
I=
Therefore,
1
(u  3) udu
2 
1
(u u  3 u )du
2 
3
1
1
=  (u 2  3u 2 )du
2
1 2 52
2 3
=
( u  3 u 2 )+c
2 5
3
3
1 52
=
u  u 2 +c
5
5
3
1
= (x 2  3) 2  (x 2  3) 2 +c
5
I =
Characteristic examples of this case are integrals of the form

polynomial
ax  b
dx , 
where we let u=ax+b
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EXAMPLE 3
I= 
Find
x2
dx
x2
Solution
Let
u=x+2 (so that x=u-2)
Then
du
= 1  dx=du
dx
Thus,
I= 
(u- 2) 2
x2
u 2  4u  4
du = 
du = 
du
u
u
u
4
u2
)du =
 4u  4lnu  c
u
2
=  (u  4 
=
(x  2) 2
 4(x  2)  4ln(x  2)  c
2
Another popular substitution of this kind is u=ex due to its simple
derivative. When you see rational expressions containing ex, think of
this substitution!
EXAMPLE 4
I 1= 
Find
ex
dx
e 2x  4
I 2= 
e2x
dx
ex  4
Solution
For both integrals:
let
u= ex,
then
du
du
= ex  dx= x
dx
e
Then
I 1= 
u dx
du
1
ex
1
u


arctan

c

arctan
c
2
2
2
u2  4 ex  u2  4 2
I 2= 
u 2 dx
u
u  4- 4
4

du  
du   1 
du
x
u4 e
u4
u4
u4
while
 u  4ln | u  4 | c  e x  4ln | e x  4 | c
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 CASE 2: Let x=(expression of u) [instead of u=(expression of x)]
In this case, the substitution is not very obvious and it is usually
given in an exam!
We will see two characteristic substitutions of this kind.
EXAMPLE 5
For I= 
dx
x
, we use the substitution x=2tanu (so u= arctan )
2
x 4
2
dx
= 2sec 2 u  dx= 2sec 2 u du
du
We have,
Thus,
I= 
2sec 2 udu
1 sec 2 udu
1 sec 2 udu 1
=  du


2
4tan 2 u  4 2  tan 2 u  1 2  sec 2 u
1
1
x
u  c = arctan  c
2
2
2
=
EXAMPLE 6
dx
x
, we use the substitution x=2sinu (so u  arcsin )
2
4- x 2
For I= 
dx
=2cosu  dx=2cosudu
du
We have,
Thus,
2cosudu
I =
2
4 - 4sin u
= u  c = arcsin

2
cosudu
cosu
= du

2  1- sin 2 u  cosu 
x
c
2
In general, the two formulas (of the formula booklet)
1
1
x
1
 a  x dx  a arctan a  c
2
2
x
 a  x dx  arcsin a  c
2
2
can be shown by suing the substitutions x  atanu and x  asinu
respectively.
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We can see these two substitutions in similar cases:
see expression
use substitution
a2  x 2
x  atanθ
2
a x
2
x  asinθ
EXAMPLE 7
Find I=  16 - x 2 dx , by using the substitution x=4sinθ
Solution.
We have,
Thus,
dx
=4cosθ  dx=4cosθdθ
du
I=  16 - 16sin 2 θ 4cosθ d θ  16  1 - sin 2 θ cos θdθ  16  cos 2 θdθ
We use the double angle identity
cos 2 θ 
cos2θ  1
2
Thus
 sin2θ

I  8 (cos2θ  1)dθ =8 
 θ  +c =4sin2θ+8θ +c
 2

But
θ = arcsin
x
4
and
4sin2θ = 8sinθcosθ = 2xcosθ
= 2x 1- sin 2 θ = 2x 1 Therefore,
I=
x
2
x2 x
=
16 2
16- x 2 +8arcsin
110
16- x 2
x
+c
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TOPIC 5: CALCULUS
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Look at the three integrals below. Although of very similar form,
they have completely different results. The difference lies in the fact
that the three quadratics involved have two, exactly one, or no real
roots respectively.
EXAMPLE 8
Consider
I1  
2
dx
x  4x  3
I2  
2
2
dx
x  4x  4
2
I3  
2
dx
x  4x  5
2
Hints:
For I1 it is given that
2
1
1


.
x  4x  3 x  3 x  1
2
[easy to confirm!]
For I2 notice that the quadratic is a perfect square.
For I3 use the vertex form of the quadratic: x 2  4x  5  ( x - 2) 2  1
Solution
1 
 1
I1   

dx  ln x  3  ln x  1  c
 x 3 x 1
I2  
2
2
dx   2(x - 2)- 2 dx  
c
2
x- 2
(x - 2)
I3  
2
dx  2arctan(x  2)  c
(x - 2) 2  1
EXAMPLE 9
Use the vertex form of - x 2  4x  3 to find
I
2
2
- x  4x  3
dx
Solution
The vertex form is –(x-2)2+1.
Thus
I
2
1- (x - 2) 2
dx  2arcsin(x  2)  c
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5.17 INTEGRATION BY PARTS (for HL)
 DISCUSSION
In this paragraph we study integrals of the form
I=  (f  g )dx
Let’s make it clear from the very beginning that there is no
“product rule” for integrals. It is sometimes very difficult, and
more often impossible, to find the indefinite integral of a product.
However, we exploit the product rule for differentiation
(f  g)  f   g  f  g 
which gives
f   g  (f  g)  f  g 
If we apply integration on both sides we obtain the so-called
Integration by parts formula
 f ' gdx =f.g-  f  g' dx
This formula does not give an answer for any product but in some
cases the integral in the RHS is much easier than the original and
thus we obtain a result.
 THE METHOD
Consider
I=  xe x dx
Since (ex)΄=ex this integral can be expressed as
I=  (e x )xdx
The integration by parts formula gives
I = exx-  e x x' dx
= exx-  e x dx = xex- ex+c
[you may easily confirm that the derivative of the result gives xex]
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In fact, the process is even quicker. We integrate one of the factors
and then we differentiate the other factor as follows
derivative
 f '  gdx = f . g -  f  g' dx
integral
For example, in I=  xe x dx we integrate e x and then differentiate x
x
x
 xe dx = x ex -  e dx
and thus the final result is
xex-ex+c
Notice
If we try to integrate the other factor, that is x , and then
differentiate e x , we obtain
x
 xe dx 
x2 x
x2 x
e 
e dx
2
2
The result of course is not wrong, but it is not practical! The second
integral is worse than the original!
But how do we choose the function we integrate first?
We follow the priority list below
Priority (for integration)
1. e x , sinx, cosx
2. xn (or polynomials)
3. lnx, arctanx, arcsinx, arccosx
In the following easy examples we indicate by a double underscore
the factor we integrate and by a single underscore the function we
differentiate.
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EXAMPLE 1
(a)
 xcosxdx  xsinx   sinxdx  xsinx  cosx  c
(b)
 e (2x  5)dx = ex( 2x  5 )-  2e dx
x
x
= ex(2x+5)- 2ex + c
= 2xex + 3ex + c
(a)
2
x
2
x
2
x
2
x
 x e dx = x ex-  2xe dx
= x ex- 2  x e dx
= x ex- 2[ xe   e dx]
[we repeat the rule]
x
= x 2 ex- 2xe x +2 e x +c
(b)
1
2
1
3
1
 x lnxdx = 3 x3 lnx - 3  x x dx
1 3
1
x lnxx 2 dx

3
3
1 3
1 3
=
x lnxx +c
3
9
=
(e)
I=  lnxdx = ?
Well, we do not see any product here! But this can be written as
 1  lnxdx
and 1=x0 is one of the factors mentioned in the priority list!
The integration by parts gives
I=  1  lnxdx =xlnx -  x 
1
dx = xlnx-  dx = xlnx-x+c
x
[ Verification: the derivative of y=xlnx-x+c is y΄=lnx+x
1
-1=lnx ]
x
In the following example we calculate the integral I   e x sinxdx .
Notice that both factors lie in the first priority. You may choose
any factor you like (I bet you will choose ex). The result is quite
interesting!
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EXAMPLE 2
Find
I =  e x sinxdx
Solution
We choose ez for integration:
I =  e x sinxdx =exsinx-  e x cosxdx

[we carry on; choose again ex]
= exsinx- e x cosx   e x sinxdx

= exsinx-excosx-  e x sinxdx
In the final expression we obtain again the original integral I, which
seems to lead to a dead-end. But in fact we have
I = exsinx+excosx- I
We solve for I and obtain
2I = exsinx+excosx
and finally
I=
e x sinx  e x cosx
+c
2
 FURTHER OBSERVATIONS
In the priority list above we may have some variations of the
factors

Instead of ex you may have eax, or any exponential ax

Instead of sinx, cosx you may have sin(ax), cos(bx)

Instead of lnx you may have any logarithm logax
EXAMPLE 3
1 2 3x 2
x e   xe3x dx
3
3
1 2 3x 2  1 3x 1 3x 
 x e   xe   e dx 
3
3 3
3

1 2 3x 2 3x
2 3x
 x e  xe 
e c
3
9
27
I =  x 2 e3x dx 
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 CASES OF INTEGRATION BY PARTS
General Form
Examples
nZ+, a,bR
In   x n e x dx
n
Theoretical Questions
x
3 x
2
 x e dx ,  x e 2 dx
ax
I n,a   x e dx
In   x n cosxdx
 x cosxdx
2
n
In   x sinxdx
In,a   x n cos(ax)dx
Express In in terms of In-1
Hence find I 0 , I1 , I 2 ,…
Express In in terms of In- 2
2
 x cos3xdx
n
In,a   x sin(ax)dx
dx
 x lnxdx ,  lnx
x
Ia   x alnxdx
5
Ia,b   e ax sin(bx)dx
 e sin2xdx
Find a general formula for Ia,b
2
Express In in terms of In- 2
-x
ax
Ia,b   e cos(bx)dx
In   cos n xdx
3
 cos xdx ,  cos xdx
n
In   sin xdx
Ia,b   sin(ax)cos(bx)dx
In   x n arctanxdx
Find a general formula for Ia
Hence find I 2 , I 4 and I3 , I5
Find a general formula for Ia,b
 sin2xcos3xdx
 arctanxdx ,  xarctanxdx ,  x arctanxdx
2
n
In   x arcsinxdx
2
 arcsinxdx ,  x arcsinxdx
In   x n arccosxdx
In   lnx  dx
n
 (lnx) dx ,  (lnx) dx
2
3
EXAMPLE 4
If In   cos n xdx , express In in terms of In- 2
In   cosxcos n-1 xdx  sinxcos n 1 x  (n  1) sin 2 xcos n- 2 xdx
 sinxcos n 1 x  (n  1)  (1- cos 2 x)cos n- 2 xdx
 sinxcos n 1 x  (n  1)  (cos n- 2 x - cos n x)dx
 sinxcos n 1 x  (n  1)In- 2  (n  1)In
 In +(n-1) In  sinxcos n 1 x  (n  1)In- 2
 n In  sinxcos n 1 x  (n  1)In- 2
Thus
In 
1
n 1
sinxcos n 1 x 
In- 2
n
n
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 INTEGRATION BY PARTS FOR DEFINITE INTEGRALS
Again, as in the case of substitution, it would be safe to find the
indefinite integral first and then proceed to the definite integral!
Otherwise the integration by parts for definite integrals takes the
form
 f '  gdx = f  g -  f  g' dx
b
b
b
a
a
a
EXAMPLE 5
Find
2
I=  e x (2x  5)dx
0
Solution
Method A: we find the indefinite integral first
x
 e (2x  5)dx =2xex+3ex+c
Therefore,

[ example 1(b) above ]

I= 2xe x  3e x 02 =(4e2+3e2)-(0+3)=7e2-3
Method B: we keep the boundaries
2


2
2
I =  e x (2x  5)dx = e x  (2x  5) 0 -  e x  2dx
0

0

2
= (9e2-5) - 2e x 0
=(9e2-5)- (2e2-2) = 7e2-3
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5.18 FURTHER AREAS BETWEEN CURVES - VOLUMES (for HL)
Consider the curve
y x
We found in paragraph 5.10 that
9
A
0
x dx  18
9
B   (3- x )dx  9
and
0
An alternative way to measure the area is to move in y-axis:

consider the functions as x in terms of y

let y move in y-axis (instead of x in x-axis).
Thus we have two formulas for the area
about x-axis
about y-axis
b
b
Area=  ydx
Area=  xdy
a
a
In our example above,
y  x  x  y2
and y ranges between 0 and 3:
3
 y3 
B=  y dy     9  0  9
0
 3 0
3
2
3

y3 
A   (9- y )dy   9y 
  (27  9)  0  18
0
3 0

3
2
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NOTICE
If y=f(x)  x=f-1(y) the alternative formula
b
b
a
a
Area=  xdy =  f -1 (y)dy
can be written as
b
Area=  f -1 (x)dx
a
In other words, we can write the inverse function in terms of y
(followed by dy) or in terms of x as usual (followed by dx).
In our example above
3
2
 y dy
and
0
3
2
 x dx
0
are exactly the same.
EXAMPLE 1
Find the area among y=lnx, x-axis and the line x=e.
About x-axis:
A   lnxdx  xlnx  x 1  (elne  e)  (1ln1  1)  1
e
e
1
About y-axis:
y=lnx x=ey
Thus
1


1
A   (e- e y )dy  ey  e y 0  (e  e)  (0  1)  1
0
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 VOLUME OF REVOLUTION
a
b
If we rotate the graph of y=f(x), 3600 about the x-axis, we will
obtain a 3-D shape as follows:
The volume of this shape is directly given by (the proof is omitted!)
b
V= π  y 2 dx
a
If we have two curves
y1=f1(x)
and
y2=f2(x)
with f1(x) ≥ f2(x) ≥ 0
the solid generated when the region between the two curves is
revolved 3600 about x-axis is given by
b
2
2
V= π  (y1 - y2 )dy
a
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EXAMPLE 2
Consider the segment of the straight line
y=
1
x , where 0  x  4 .
4
Find the volume of the cone generated by a 3600 rotation of this
segment.
b
4
a
0
V = π  y 2 dx = π  (
1 2
x) dx
4
4
π 4 2
π x3 
=
x
dx
=
16 0
16  3  0
=
π 4 3 4π
=
16 3
3
[Notice that the known formula for the volume of the cone of
height h=4 and radius r=1 gives:
1
4π
V= π r 2 h =
3
3
as expected!
If the rotation takes place about the vertical axis (y-axis), then we
solve y=f(x) for x, (hence x=f-1(y)), and the formula now is
b
V= π  x 2 dy
a
Notice that the solids generated when a curve is rotated in x-axis
or in y-axis are completely different.
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EXAMPLE 3
2
If we rotate the parabola y=x2 about the y-axis we obtain the 3D
shape above! Find the volume if the height is 4.
We solve for x, so that x= y .
The volume is
4
 y2 
V= π  x dy = π  x dy = π  ydy =π   =8π
a
0
0
 2 0
b
4
2
2
4
EXAMPLE 4
Consider the region between the curves y=2x2 and y=2x. The two
curves intersect at x=0 and x=1.
If the region is rotated 2π about x-axis, the volume generated is
1
x3 x5 
 1 1  8π
V= π  (4x - 4x )dy = 4π 

 =4π    
0
5 0
 3 5  15
3
1
2
4
If the region is rotated 2π about y-axis, the volume generated is
2
 y y2 
 y 2 y3 
2 π

dy = π 
V= π   

 =π 1   

0 2
3 3
4 


 4 12  0
2
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5.19 DIFFERENTIAL EQUATIONS (for HL)
You already know some simple differential equations:
Find f(x) given that
f (x)  8x 3 , with f(1)  5
Or otherwise, express y in terms of x given that
dy
dx
 8x 3 , with y=5 when x=1
Clearly, by integration we find
y  2x 4  c
The condition y=5 when x=1 (or otherwise y(1)=5) gives
c=3
Thus the final answer is
y  2x 4  3
The terminology we use is as follows
dy
 8x 3
differential equation
y(1)=5
boundary condition
dx
y  2x 4  c
general solution
y  2x 4  3
particular solution
Well, a differential equation relates a function y=f(x) with its
derivatives. Thus, it involves
x, y, y , y , etc.
Our task is to find the functions y=f(x) that satisfy this relation.
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For example,
xy  2y- y(x  2)  x 2  x
is a differential equation and
y  ex  x
is a particular solution, since
y  e x  1
y  e x
and
LHS = x(e x  1)  2e x  (e x  x)(x  2)
= xe x  x  2e x  xe x  2e x  x 2  2x
= x  x2
= RHS
Perhaps there are more solutions! As we said, our task is to find
the whole family of functions y=f(x) that satisfy the differential
equation.
This is a 2nd order differential equation as it involves the derivatives
up to y .
In general, the order of a D.E. is the number of the highest
derivative in the equation.
Here we only deal with 1st order D.E. that is our equations involve
only
x, y and
dy
dx
We will investigate only 3 particular cases of 1st order D.E.

D.E. of separable variables

Homogeneous D.E.

Linear D.E. (with integrating factor)
We will also present a numerical method (approximation) for D.E.

Euler’s method
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TOPIC 5: CALCULUS
Christos Nikolaidis
 D.E. OF SEPARABLE VARIABLES
A differential equation is of separable variables if we can separate
x’s from y’s and express the equation in the form
f(y)dy=g(x)dx
Then we integrate both parts and get the result
 f(y)dy   g(x)x
EXAMPLE 1
Solve the D.E.
dy
dx
=4xy2, given that y(1)=2.
Solution
We separate the variables:
dy
dx
=4xy2 
dy
y2
 
=4xdx
dy
y2
=  4xdx
1
= 2x 2  c
y
1
 y =
2x 2  c
 
[general solution]
The boundary condition gives
1
5
y(1)=2  
 2  -1=4+2c  c=2c
2
Therefore,
y
or equivalently
y
1
5
2
2x 2 
2
5 - 4x 2
Even the simple example
dy
dx
[particular solution]
 8x 3 in the introduction can be seen
as a D.E. of separable variables:
dy
dx
 8x 3  dy=8x3dx   dy = 8x 3 dx  y  2x 4  c
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TOPIC 5: CALCULUS
Christos Nikolaidis
The evaluation of the constant c can be done in an earlier step.
EXAMPLE 2
Solve the D.E.
dy
dx
 xy 2  x , given that y(0)=1.
Solution
dy
dx
 xy 2  x 

dy
 x(y 2  1)
dx
dy
2
y 1
 
 xdx
dy
2
y 1
  xdx
 arctany 
x2
c
2
Since y(0)=1,
arctan1  c  c 
[general solution]
π
4
Therefore,
x2 π

2 4
 x2 π 
 y  tan 
 
 2 4
arctany 
[particular solution]
A common problem in this category is to find a population P where
the rate of increase (or decrease) is proportional to the population
itself:
By separating variables
dP
 kP
dt
dP
dP
 kdt  
  kdt  lnP  kt  c  P  e kt c  e kt e c
P
P
By setting P0  e c (initial population) we obtain
P  P0 e kt
that is an exponential growth for the population in terms of time.
126
TOPIC 5: CALCULUS
Christos Nikolaidis
 HOMOGENEOUS D.E.
A differential equation is called homogeneous if it can take the form
dy
y
= F  
dx
x
y
[I.e. the RHS is a function of
x
]
It is not so difficult to recognize the homogeneous D.E.
For example,
dy
dx
dy
dx
dy
dx
dy
dx

x 2  xy  y 2

x 2 y  y3


x2
dy
y y
 1    
dx
x x
becomes
dy
becomes
x3
x 2  y2
dx
dy
becomes
xy
x 3  2y 3
x 2 y - 3x 3

dx
y
  
x x
3
x 2  y2

dy
becomes
y
2
dx
x2
xy
x2
x 3  2y 3

x3
x 2 y - 3x 3
x3

 y
1   
dy
x

y
dx
2
x
y
1  2 
dy
x


y
dx
-3
x
Then we use the substitution
v
dy
dx
and the D.E.
y
x
 y  xv
=v+x
dv
dx
[by using product rule]
dy
y
= F   takes the form
dx
x
v+x
dv
=F(v)
dx
which is always a D.E. of separable variables (v in terms of x)
[can you see why?].
127
3
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 3
Find the general solution of the D.E.
x2
dy
dx
 x 2  xy  y 2 .
Solution
x
Let
Then
dy
2
dx
v
dy
dx
y
x
2
 x  xy  y
2

dy
dx

x 2  xy  y 2
x2
dy
y y
 1    

dx
x x
 y  vx
=v+x
dv
dx
Thus
v+x
dv
=1-v+v2
dx
 x
dv
=1-2v+v2
dx
 x
dv
=(v-1)2
dx

dv
dx
=
2
x
(v - 1)
 
dv
dx
=
2
x
(v - 1)
 
1
= ln x +c
v- 1
 v- 1 = 
 v =1 
Finally, since v 
y
x
1
ln x  c
1
ln x  c
,

y
1
1 
=1 
 y= x  1 

ln x  c
x
ln x  c 

128
2
TOPIC 5: CALCULUS
Christos Nikolaidis
 1st ORDER LINEAR D.E. (WITH INTEGRATING FACTOR)
The last category contains the D.E. of the form
dy
dx
 P(x)y  Q(x)
where P(x) and Q(x) are functions of x only.
Methodology:

we spot P(x) and Q(x)

we find the so called integrating factor   e 

It holds

We calculate the integral in the RHS and solve for y
P(x)dx
(ignore +c)
y   Qdx
We will demonstrate the procedure by using an easy example and
then we will explain why this method works!
EXAMPLE 4
Find the general solution of the D.E.
dy 2
 y  5x 2 .
dx x
Solution
P(x) 
2
x
and
Q(x)  5x 2 .
The integrating factor is
  e
ignore +c
P(x)dx
2
 dx
 e x  e 2lnx  x 2
Then
y   Qdx  x2y   5x 4 dx  x 5  c
Therefore
y  x3 
129
c
x2
TOPIC 5: CALCULUS
Christos Nikolaidis
Explanation:
Differential equation:
dx
 P(x)y  Q(x)
  e
Integrating factor:
Notice that
dy
P(x)dx
d  d   P(x)dx    P(x)dx 

e
  e
P(x)  P(x)
dx dx 
 

If we multiply the D.E. by  we obtain

dy
 P(x)y  Q(x)
dx
But the LHS is the derivative of the product y [can you see why?]
Thus
d
(y)  Q(x)
dx
 y   Q(x)dx
Provided that the last integral is easy to find we can solve for y
and obtain the result.
Based on this explanation, let us provide a more analytical solution
for the D.E. in Example 4:
dy 2
 y  5x 2 .
dx x
The integrating factor is
2
P(x)dx
 dx
  e
 e x  e 2lnx  x 2
We multiply the equation by x2:
x2
dy
 2xy  5x 4 
dx
d
(x 2 y)  5x 4
dx
Thus
x 2 y   5x 4 dx  x 5  c
130

y  x3 
c
x2
TOPIC 5: CALCULUS
Christos Nikolaidis
 EULER’s METHOD: A NUMERICAL SOLUTION
Consider the D.E.
dy
 f(x, y) ,
dx
y(x0)=y0
where f(x,y) is an expression in terms of x and y.
By using a step h, we find consecutive points
(x0,y0), (x1,y1), (x2,y2), …
that approximate the solution.
We complete a table of the form
n
xn
yn
0
x0
y0
1
2
by using the recursive relations
xn+1 = xn + h
yn+1 = yn + hf(xn,yn)
We will demonstrate the procedure by using the D.E. of example 2
again and then we will explain the method!
EXAMPLE 5
dy
dx
 xy 2  x ,
with y(0)=1.
Find an approximation of y(1) using step h=0.2
Solution
The relation xn+1 = xn + h = xn + 0.2 gives directly the column of xn
n
xn
yn
0
0
1
1
0.2
1
2
0.4
2
0.6
4
0.8
5
1
131
TOPIC 5: CALCULUS
Christos Nikolaidis
The relation
yn+1 = yn + hf(xn,yn) = yn + 0.2( x n yn2  x n )
gives
y1 = y0 + 0.2( x 0 y 02  x 0 ) =1
y2 = y1 + 0.2( x 1 y12  x 1 ) =1.08
y3 = y2 + 0.2( x 2 y22  x 2 ) =
etc
We finally obtain
n
xn
yn
0
0
1
1
0.2
1
2
0.4
1.08
2
0.6
1.25331
4
0.8
1.56181
5
1
2.11209
Hence
for x=1, y  2.11209
Notice

We can easily obtain the table above by using recursion in your
GDC.
For Casio FX we use
MENU
RECURSION
SET[F5]
Start=0,
End =100 (or more),
a0=0,
b 0= 1
EXIT
an+1 = an + 0.2
bn+1 = bn + 0.2( an bn2  an )
EXE
132
TOPIC 5: CALCULUS

Christos Nikolaidis
 x2 π 
The exact solution of the D.E. is y  tan 
  [see example 2]
2
4

Thus for x=1, the actual value of y is
1 π
y= tan     3.40822
2 4
The approximation y  2.11209 we found above is not great!
However, we can improve the result by reducing the step value.
For h=0.01, (thus we need 100 steps) the GDC gives
for x=1, y  3.26945
which is much better.
Explanation of the method:
dy
 f(x, y) ,
dx
y(x0)=y0
actual value
approximation
The tangent line at (x0,y0) of the unknown solution curve is
y  y 0  m(x  x 0 )
But m 
dy
 f(x 0 , y0 ) , so
dx (x ,y )
0
0
y  y0  f(x 0 , y0 )(x  x 0 )
 y  y0  (x - x 0 )f(x 0 , y0 )
For x=x1, we let h=x1-x0 and approximate the actual value of y by
y1  y0  hf(x 0 , y0 )
Similarly we get
y2  y1  hf(x 1 , y1 )
and so on.
133
TOPIC 5: CALCULUS
Christos Nikolaidis
5.20 MACLAURIN SERIES – EXTN OF BINOMIAL THM (for HL)
Consider the infinite geometric series
1  x  x 2  x3 …
We know that the series converges for -1<x<1 and the result is
S 
1
1x
In this paragraph we have the opposite task:
1
, and we wish to express it
We are given a function, say f(x) 
1x
as an infinite series (which is called power series) of the form
a0  a1 x  a2 x 2  a3 x 3  …
(the power series looks like a polynomial of “infinite degree”).
 THE MACLAURIN SERIES
Suppose that a function f(x) has derivatives of every order near 0.
Then f(x) can be expressed as a power series as follows
f (0) 2 f (0) 3
x 
x …
2!
3!
This is known as Maclaurin series of the function.
1
For example, if f(x) 
then
1x
f(x)  f(0)  f (0)x 
f (n) (x)
f (n) (0)
f(x)  (1  x)-1
1
f (x)  (1  x)-2
1
f (x)  2(1  x)-3
2
f (x)  6(1  x)-4
3!
etc
and thus
f(x)  1  x  x 2  x 3  …
It is in fact the geometric series in the beginning. Amazing, isn’t it?
134
TOPIC 5: CALCULUS
Christos Nikolaidis
Explanation of the Maclaurin series:
We wish to express f(x) as
f(x)  a 0  a1 x  a 2 x 2  a3 x 3  a 4 x 4  …
Then
f (x)  a1  2a 2 x  3a3 x 2  4a 4 x 3 …
f (x)  2a 2  3! a3 x  (3)(4)a 4 x 2  …
f (x)  3! a3  4! a 4 x  …
etc
Therefore,
f(0)  a 0
 a 0  f(0)
f (0)  a1
 a1  f (0)
f (0)
2
f (0)
f (0)  3!a 2  a3 
3!
f (0)  2a 2  a 2 
etc
In general
an 
f (n) (0)
n!
In fact, the partial sums of the Maclaurin series give good
approximations of f(x) near x=0:
a 0  a1 x
is the tangent line of f(x) at x=0
a0  a1 x  a2 x 2
2
a0  a1 x  a2 x  a3 x
For f(x) 
is the “best” quadratic that approximates f(x)
3
is the “best” cubic that approximates f(x)
1
(black curve) look at the approximations below:
1x
y  1  x [tangent]
y  1  x  x2
135
y  1  x  x 2  x3
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 1
Find the Maclaurin series of the function f(x)=sinx up to the term in x5
Solution
f (n) (x)
f (n) (0)
f(x) = sinx
0
f (x)  cosx
1
f (x)  -sinx
0
f (x)  -cosx
-1
(4)
(x)  sinx
0
f (5) (x)  cosx
1
f
and thus
f(x)  f(0)  f (0)x 
f (0) 2 f (0) 3
x 
x …
2!
3!
gives
sinx  x 
x3 x5

…
3! 5!
Notice: Use your GDC to compare the graph of sinx with the graphs
of the partial sums
x
The result is amazing!
x3
x3 x5
, x
,

3!
3! 5!
Similarly, we can obtain
ex  1  x 
cosx  1 
x2 x3

…
2! 3!
x2 x4

…
2! 4!
x 2 x3
ln(1  x)  x 

…
2
3
arctanx  x 
x3 x5

…
3
5
136
etc
TOPIC 5: CALCULUS
Christos Nikolaidis
An alternative way to obtain a Maclaurin series is to modify or
combine appropriately the already known series. We can add,
multiply, differentiate or integrate series and generate new series.
EXAMPLE 2
Find the Maclaurin series of the function f(x)  e x
2
Solution
Since
ex  1  x 
x2 x3

…
2! 3!
we can substitute x by x2 and obtain
2
ex  1  x 2 
x 4 x6

…
2! 3!
EXAMPLE 3
Find the Maclaurin series of the function f(x)  e x sinx
Solution
We can multiply the series for e x and sinx.



x 2 x3
x3 x5
e x sinx   1  x 

 …  x 

 …
2! 3!
3! 5!



We can find gradually the constant term, the terms in x, the terms
in x2 etc:
e x sinx  x  x 2 
 x  x2 
x3 x3 x4 x 4



…
3! 2! 3! 3!
x3
…
3
[there is no x4]
Notice also

if we differentiate the series for sinx (term by term) we obtain
the series of cosx.

If we differentiate the series for ex we obtain ex itself.
137
TOPIC 5: CALCULUS
Christos Nikolaidis
In general, we are able to differentiate a series term by term.
EXAMPLE 4
Find the Maclaurin series of the function f(x) 
1
x 1
2
Solution
arctanx  x 
Since (arctanx) 
1
x 1
x3 x5

…
3
5
2



1
x3 x5
 x 

 …  1  x 2  x 4  …
3
5
x 2  1 

We can also integrate term by term.
EXAMPLE 5
Find the Maclaurin series of cosx by integrating the series of
sinx  x 
x3 x5

…
3! 5!
Solution
(a) By using indefinite integrals:
 x2 x4



x3 x5




sinxdx

x



…
dx
cosx


 …  c

  3! 5! 
 2! 4!

For x=0 we obtain -1=c. Thus
 x2 x4

x2 x4
- cosx  

 …  1  cosx  1 

…
2! 4!
 2! 4!

(b) By using definite integrals from 0 to x:
x
x
x



x2 x4
x3 x5
x



sinxdx

x



…
dx

cosx



 …
0

0
0  3! 5! 
 2! 4!
0
 x2 x4

x2 x4
 - cosx  1  

 …  0  cosx  1 

…
2! 4!
 2! 4!

138
TOPIC 5: CALCULUS
Christos Nikolaidis
 DIFFERENTIAL EQUATIONS AND MACLAURIN SERIES
Consider a differential equation of the form
dy
 F(x, y)
dx
with boundary condition
y(0)=y0
The analytical solution is not always easy or sometimes not possible
at all. However, we can easily find the Maclaurin series of the
solution y=f(x) (and thus a good approximation of the solution).
f(x)  f(0)  f (0)x 
But,
f (0) 2 f (0) 3
x 
x …
2!
3!
[by the boundary condition]
f(0)= y0
f (0) =
dy
 F(0, a)
dx x  0
[by the D.E. itself]
y a
Implicit differentiation on
dy
dx
gives
d2y
dx 2
and thus f (0) , and so on.
EXAMPLE 6
Find the Maclaurin series up to x2 for the solution of the D.E.
dy
dx
 x 2  y2
with y=3 when x=0
Solution
Clearly
f(0)=3
f (0) =
dy
dx x  0
 0 2  32  9
y 3
d2y
dy
dx
dx
 2x  2y
2
, thus
f (0) =
d2y
dx 2 x  0
 2  0  2  3  9  54
y a
Therefore
54 2
x  …  3  9x  27x 2  …
2!
d2y
Notice. We can also express
in terms of x and y only:
dx 2
d2y
dy
 2x  2y
thus f (0) =54
 2x  2y(x 2  y 2 ) ,
2
dx
dx
y  3  9x 
139
TOPIC 5: CALCULUS
Christos Nikolaidis
 THE EXTENTION OF THE BINOMIAL THEOREM
Remember that the binomial theorem gives
 n  n
n
n
(1  x) n =   +   x +   x 2  ⋯    x n
 0  1
2
n
If we expand more the coefficients we obtain
(1  x) n = 1 +nx+
n(n - 1) 2 n(n - 1)(n - 2) 3
x +
x ⋯+ xn
2
3!
This version allows us to use negative or even fractional values for
index n.
Thus for example,
(1  x)-n = 1- nx +
- n(-n - 1) 2
- n(-n - 1)(-n - 2) 3
x +
x ⋯
2
3!
but now we have infinitely many terms.
This is an alternative way to obtain some infinite power series.
For example
Similarly,
1
 (1  x) 1  1  x  x 2  x 3  …
1 x
1
 (1  x) 1  1  x  x 2  x 3  …
1x
EXAMPLE 7
Find the Maclaurin series up to x4 for f(x) 
x
(1  x)3
Solution
(-3)(-4) 2 (-3)(-4)(-5) 3


f(x)  x(1  x)-3  x  1  3x 
x 
x  ⋯
2
3!


and finally
f(x)  x  3x 2  6x 3  10x 4  ⋯
140
TOPIC 5: CALCULUS
Christos Nikolaidis
EXAMPLE 8
Find the Maclaurin series up to x3 for f(x)  1  x by using the
extension of the binomial theorem.
Solution
The formula
(1  x) n = 1 + nx +
gives
f(x)  (1  x)
and finally
1
2
n(n - 1) 2 n(n - 1)(n - 2) 3
x +
x ⋯
2
3!
1 1
1 1
3
(- )
(- )(- )
1
2 x3 ⋯
= 1 + x + 2 2 x2 + 2 2
2
2
6
f(x)  1 +
1
1
1 3
x- x2 +
x ⋯
2
8
16
EXAMPLE 9
Find the Maclaurin series up to x2 for f(x)  2x  3
2
by using the
extension of the binomial theorem.
Solution
f(x)   3  2x 
2
 
2x  
 3  1 

3  
 
2
1
2x 
 1 

9
3 
The formula
(1  x) n = 1 + nx +
gives
f(x) 
n(n - 1) 2
x ⋯
2
2

1 
2x (-2)(-3)  2x 
12

 ⋯



9 
3
2
 3 

and finally
f(x) 
1 4
4 2

x
x ⋯
9 27
27
141
2