Class notes

MAM1012S- Mathematics 11

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Course
(MAM1012S)

Institution
University Of Cape Town (UCT)

ALL THE NOTES YOU WILL NEED TO PASS THE EXAM. First-year second-semester mathematics notes, all chapters start to finish. Notes made from the EDU lecture slides, textbook, lecture examples and lecturer talking points thus making them in-depth and comprehensive.

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Uploaded on November 3, 2023
File latest updated on November 4, 2023
Number of pages 77
Written in 2022/2023
Type Class notes
Professor(s) Na
Contains All classes

mathematics

Institution
University of Cape Town (UCT)
Course
(MAM1012S)

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finding the average/ mean of sets -

add all the values $ divide by the amount of no .
U added

averages :

① { 2 7 3 5 }
avg / mean of , , ,
,

a

◦
: 17/5 = 3 . 4

② R 54
avg price of R42
,
R 50
,

◦
: 421-501-54 = R 48 67
,
3

continuous function
^
y areas of roam -

wax

↳ under the
ya area of rectangle = area curve

, under
area curve
ya c
=
✗

'
under curve
avg Val =
area

C
b
c

avg Val = ¥a f a
fix) doc

example
① Average value
7
of fact = od over [ 3,7]

§ = ÷ JOE
3
doc
area under

¥?:::
=
Ya [ x% ]} curve =

=
'
it [% -

3% ] rectangle in

.
=
79/3 = 26 ,
3
S

,2.) savings account , 3% interest compounded continually

At year end bonus = 1% of average balance in account during the year -

Deposit R10 000 at start, what is bonus at end of year 1?

" °"

s cont .

comp .
A- (f) =
10000 @
b

avg balance =
A- =¥a
'
J Alt) dt
a

= ¥0 J 10000 @
" ° "

dt

0

[ t.org ];
◦ ' 03£
=
10000 @

10151,51
=

Bonus = It × 10151,51

=
R 101,52

, moving averages (discrete & continuous)

when we have a sequence of data points Iz Iz the values are likely to
,
JC ,
, ,
-
. . ✗ 20

jump up and down. This makes it difficult to discern if there is a trend
•
: we use moving averages to:

- smooth it irregularities
- show trends
NB! the higher the n the smoother the plot but the greater the lag
EXAMPLE:

1.) find the 5 day moving average

SO! to move the average we construct a new sequence of data by taking the
average of the first 5 points

:⊖
÷/É::::
Moving
22.2 22.8 23.4 24.4 23,6 23,6

20

i. e. f- (5) =É[¥ ,
Xi then f- (6) =É%=z0Ci . . .
f- ( 201=5 [ i. *
Oci

: the n-unit moving average of a function:
✗ → endpoint

fix tnffcxldx-

_

↓ 2C n-
n

OE

an → begin point

ifeng.tn

, "
avg on
[a. b)
Gee subint points
ya
f. f ca
=
+ Éox )
f⇐ ,

f- ≈

(ft tfz
600C
t - - -

f- 6
] ox

a
'

b ¥
t.floxtfzoxt.r.tt# n
b- a

§ ≈ [
in
fi ox
↳ reimann sum !!

b- a

¥4 fi
9

As
§ S fix> dx
° "
n →
a .
=
¥ . =
↳

b- a
b- a

Example :

f
3
① C) ✗ x2 30C 3 unit
moving
+
avg
= -

,

x

§ =
↳
f ( Est E
'
-
3-c) It
x -3

"

43 [ Yat ↳ % ]
" '
=
+ t 3- t
x 3
-

= (K2 I
"
t Ya 2C
} _

tz x2 ) -

( Hz ( x 3) -
"
t Yq (x 3)-
3

hex-312 )
'
-

Ya (
f-
}
↳c) =
ax _

14 x2 + 122C +
3)

↳

Nt "

,continuous income streams
"
FU =P Vert
compounding !!
→

RK)
↳ revenue
rand
.
per day ,
t in
days

———-formula ————————————————
O

total revenue
b

total rev = TV = f Rtt ) dt

↳
a

total income over [ a. b ]

•
future value (incl interest)
'

future value = FV = fa
R (E) er
#"
dt

0
present value
"

[
↑

Puerto $
a) "
Rctjrlb
- -

FU =
↳
in
yrs
FU =
dt
a

f
"" ' "" '
-
:
Pver =
Rlt) er dt
a
b

f R (f) er
1
" " -

PU =
de e-
' b- a)
.

a

[
re rb + ra

R (f) erb
- -

=
dt
:

f R / Her
""
dt
=

a

present value = PV = ! a
R (E) er
"
de

,EXAMPLE: total value
1.) sales income from 1 nov to 31 jan, 92 days given by

R(t)= 30000 + 45t - 0,5t rand per year
total income over period?

time interval [0,92]
n - sub intervals [ tic - i
,
tic ] of length Ot

•
: revenue in any sub interval [ tic - i
,
tic ] is approx
R (tic ) -
i
Ot

Add or all sub intervals
total income = Rlto ) Ott R ( Ei ) Ott
- - .

+ R ( En - 1) Ot

92 ↳ left reiman sum

let n → a total income =

frltldt
0

92

[
=

(30001-451--0.5-4) dt

R 336659
=

EXAMPLE: future value

2.) Future value continuous income stream, deposited into account paying interest
(continuous compounding)
- ↳ Fu =P Vert

interest 5% pa, balance at end of jan?
2
$ R (t) = 3000 + 45-6 -
0,5 C-

S [0 , ]
92
,
sub int [ tic -
ist ] ,
Ot
; rev ≈ Rltk 1) -
Ot

deposited ( PV )

R
◦ ◦ s ( 92 -
tic -11/365

ACK 1)
'

-
=
of @

add for all the sub intervals

, Int earned 92 tk , days
-
:
92ggᵗg Years

↳ R (-1--0) Of @
0105 ( 92 1365 t
R( of e.
" ◦
sC92- 136% .
. .

" ◦ scaz -⇐ / 365
+ R ( € of e

92

f
5 '" "
1365dg
◦ ' ◦ -

As n → 0
,
balance Rft) @
0

EXAMPLE: present value
3.) Inga was injured and can no longer work. As a result she is awarded the present
value of income that she would have receive over the next 20 yrs. Her income at the
time of injury was $100000 /yr increasing by $5000 / yr. What will the amt she is
awarded be assuming continuous income and a 5% interest rate.

PU= [ Rfllerla -
"
dt
tin
↳
years
20
- IBP :

§( )e%◦( Hdt °
So PU f-
-

1000001-5000-1
g'
"°"
=
20T£ e-
=

=

'
◦

g. I
= ""

1. e-
5000%20 g.
=
"" "

E) e- dt
=
+ -
aos

"

5000¢20 f) °ˢᵗ
'°ˢᵗ

f
°

e-
"
1. C-
+
-

=
-
0,0s -
0,05

=

$1,792,723,35

,Improper integrals
improper integral if function is undefined in given range

definite :
↑ a
fix doc

indefinite :

f fix ) doc

0

if
" "

finite =

converges

uded/ infinite =
"
diverges
"

↳ no upper or lower limit

◦
aka no boundaries .

what if:
1.) a or b not finite

[ fix ] da := i
M→x
§ fix) dx

this allows us to evaluate the integral and then take the limit → ☒

b

I.Jfcxldx :=
ei
m→ .
. f. fixate
fix )dx : =

§ .
fcxsdx + § .
fcx) dx

= for a conveniently chosen c, provided that both integrals on RHS converge

2.) F(x) not continuous on [a, b]
↳
§
-
2
2

↳
doc
✗ not defined @ 0

integral diverges when your answer is: ✗ or -
✗

integral converges when your answer is: normal

, type 1: from a definite to an indefinite. Approaches 0
;
-
x

type 2: definite range but integrals not defined

Example :
type I Infinite limits of integration .

j ¥2
•

,
dx =
li
M→ 0
JIE ,
da

= li
→ •
1- ÷ ] ?
=
e.
m → a
⇐ -

t :-|]

=

% .
C- In + I ] .

=
I

0

similarly :

§ - A
e-
"

doe =
him →
_ •
/
M
e-
"

dx

=
I'm →
•
[ e-
"

I
=
I C- I + e-
m
)
M → -

A
=/
e- 1-01=0
= -
I t ✗

=
@

limit
'

.
'

does not exist $ integral diverges .

, It definite / improper integrals
Example :
type inter grab

integrands that become infinite

§ II doc j
•
on lo ,
i
] ¥ is defined
-
0

•
as ✗ → 0 → a
,

[# doc : lie⇐
door→ Ot
[ r

= tr
↳ ◦
+
[ 2K ]!

=
↳ + ( 2 -
2 F)

§ Éidx =
. .
_

§¥ ,
doc

=

[
2
Ézdx +
% ¥2 doc

r

=

[
2
¥ dx =
li
'
f 'xadx = I
r→◦
.
Ex ]: =

f- ÷ -

C- ÷ )]
r→ 0
,

=
liftr→o
-
-

E)

= 0 -

£ =
a

÷ limit does not exist
÷
[
-2
Édx diverges
→
if one

then
part
the
diverges
whole

thing diverges

NB ! !
3

f. ¥2
'

doc ≠ f- ¥] z
= -

§
→
¥ is
always tire . : area under
graph ≠ 've

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