H2 Mathematics Cheat Sheet by Sean Lim

8/16/2019 H2 Mathematics Cheat Sheet by Sean Lim

1/27

Page 1 of 1

BRIEF SUMMARY ON INTEGRATION AND APPLICATIONS

Techniques in integration

1) INTEGRATION BY STANDARD FORM

•

1[f ( )]f '( )[f ( )] d , 1

1

nn x x x x C n

n

!

" ! # $% !

• 1f '( )[f ( )] d ln f( ) x x x x C $ " !%

•

f ( ) f ( )f '( ) d x x x e x e C " !%

2) INTEGRATION OF RATIONAL FUNCTIONSf ( )

dg( )

x x

x%

Step 1) Proper fractions? If no, do a long division.

Step 2) Check for f ' ( ) x . If yes, integrate via standard form.

Step 3) Numerator is a constant, denominator is quadratic expression? If yes, complete the square and

use MF15.

Step 4) Numerator is a linear expression, denominator is a quadratic expression? If yes, split numerator

into the form ( )f 'a x b+ and integrate via standard form.

Step 5) If denominator is cubic, most of the time we will try partial fractions.

3) INTEGRATION BY SUBSTITUTION TECHNIQUE

Step 1) Use the substitution suggested in question and differentiate the substitution and make d x thesubject.

Step 2) Change limits if it is definite integrals

Step 3) Ensure TOTAL replacement (including integrand, limits and d x).

Step 4) After TOTAL replacement, the integrand should be simpler. If not check your working.

Step 5) Integrate and change back to original variable if it is not a definite integral.


2/27

Page 2 of 2 O a b x

y = f( x)

y = g( x) y

S

4)

INTEGRATION BY PARTS.

Step 1) u dv duv v u" $% %

Step 2) The LIATE is just a GENERAL GUIDE to determine u and dv but it does NOT hold all the

time!

Step 3) General choice of u should be an expression that can be easily removed to make integrand

simpler.

APPLICATIONS OF INTEGRATIONS

Area of regions bounded between two curves y = f( x) and y = g( x)

Draw a rough sketch of the 2 graphs and identify the region needed carefully using the descriptions given in

the question.

Area of the region S (using the x-axis)

[ ]

( ) ( )

f ( )d g( )d f ( ) g( )

where f g 0

d ,

for .

b b b

a a a

x x

x x x x x x

a

x

x b

− = −=

− ≥ ≤ ≤

! ! !

Or

Area of the region R (using y-axis)

= [ ]f ( ) g( ) dd

c

y y y−!

Volume of solid of revolution for regions bounded between two curves

Draw a rough sketch of the graph(s) concerned and identify the region needed carefully using the

descriptions given in the question. Spot keywords on the axis of rotation. The solid obtained will be very

different.

Vol of the solid formed when the region S is rotated through π2

radians about the x-axis [ ] [ ]{ }2 2

f ( ) g( ) d x

b

a x x xV = π −

!

O a b x

y = f( x) y = g( x)

y

S

O

y

x

d

c

x = g( y)

x = f ( y) R


3/27

Page 3 of 3

O

y

x

d

c

x = g( y)

x = f ( y) R

Vol of the solid formed when the region R is rotated through

π2 radians about the y-axis [ ] [ ]{ }2 2

f ( ) g( ) dyd

yc

y yV = π −!

Note: If the area computed is negative, we can put a modulus over the expression to ensure the numerical

value is positive.

Area or Volume solid of revolution for curves given

in the parametric form x = f(!) and y = g(! )

The technique or solving strategy to finding area orvolume of solid of revolution for curves given in

parametric form is quite similar to the idea of

substitution technique in definite integrals.

Draw a rough sketch of the graph(s) and identify the

region needed carefully using the descriptions given in

the question. Spot keywords on the axis of rotation.

y

xb

d

a

c

x = f(!) and y = g(!)


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F u n c t i o n s s u m m a r y

!

! " #

! $

! ! " # / ! " # !

D o m a i n

U s u a l l y g i v e n i n q u e s t i o n .

%

' (

'

) *

+

, & ) *

' & ) *

+

,

R u l e

U s u a l l y g i v e n i n q u e s t i o n .

1 .

L e t - +

! . / 0

t h e n ! " # . - 0 +

.

2 .

M a k e / t h e s

u b j e c t

3 .

S u b s t i t u t e / i n t o ! " # . - 0 +

/ .

4 .

T h e n ! " # . - 0

+

/ 1 - 2

% & .

5 .

R e p l a c e - w i t h / .

N o t e :

S o m e t i m e , a f t e r f i n d i n g t h e i n v e r s e , t h e r e a r e t w o

p o s s i b l e r u l e s . F o

r e x a m p l e ±

. W e u s e

'

t o r e j e c t o n e o f t h e m .

R

e p l a c e / i n ! . / 0 w i t h $ .

0 .

/

R a n g e

F r o m t h e s k e t c h o f t h e f u n c t i o n , w i t h t h e

h e l p o f G C .

a )

C h e c k w h e t h e r t h e r e a r e t u r n i n g

p o i n t s , a s y m p t o t e s .

b )

E n d s p o i n t s m u s t b

e i n d i c a t e d w i t h a

s o l i d d o t i f t h a t p o i

n t i s i n c l u d e d a n d

a n o p e n - c i r c l e i f i t i s n o t i n c l u d e d .

c )

T h e p o s s i b l e - v a l u e s i s t h e r a n g e o f

!

'

M

e t h o d 1

S

k e t c h ! $ .

F r o m t h e g r a p h , w e c a n f i n d

t h e r a n g e .

M

e t h o d 2

S

u b s t i t u t e % (

a s n e w d o m a i n o f f a n d

f r o m t h e

s

k e t c h o f n e w f f i n d t h e r a n g e .

F o r e x a m p l e ,

% (

+

4 5 1

6 7 .

S k e t c h

1 )

% & & ) *

+

' & ) *

2 )

% & ) * & +

' &

E x i s t

V e r t i c a l l i n e t e s t

F o r e v e r y v e r t i c a l s t r a i g

h t l i n e / +

8 ,

/ 2

' & .

I t c u t s t h e g r a p h o f ! a t e x a c t l y

o n e p o i n t . H e n c e ! e x i s t .

N o t e : W r i t e d o w n t h e v

a l u e s o f ' & .

T o s h o w ! i s n o t a f u n c t i o n , j u s t g i v e

c o u n t e r e x a m p l e .

E g . S i n c e ! . / #

0 i s u n d e f i n e d .

! i s n o t a

f u n c t i o n .

H o r i z o n t a l l i n e t e s t

F o r e v e r y h o r i z o n

t a l s t r a i g h t l i n e y

k

=

, , 8 2

% & i t

c u t s t h e g r a p h o f

f a t e x a c t l y o n e p o i n t . T h e r e f o r e

f i s o n e - o n e . ”

N o t e : W r i t e d o w n t h e v a l u e s o f % & .

T o s h o w n o t o n e -

o n e ,

j u s t g i v e a c o u n t e r e x a m p l e .

E g . S i n c e ! . / #

0 +

! .

9 0 e , t h e f u n c t i o n f i s n o t

o n e - o n e .

% (

:

'

N o t e : W r i t e d o w n t h e v a l u e s o f % & a n d

' (

A l w a y s e x i s t s a s l o n g

! " # e

x i s t !

E x t r a

% &

:

%

a

b

R g

R f g

f ( x )


15/27

Some possible advance questions.

1) f is not one-one, restrict f D .

2) Make use of the fact that the graphs of ! and ! "# are reflection of each other about the line y x= .

Sketch $ % ! &'(, $ % ! "#&'( and $ % ' on the same diagram. Then solve for ! &'( % ! "#&'(.

From the graph, we check whether that $ % ! &'( and $ % ! "#&'( intersect at the line $ % '. If they do, then solve for ! &'( % ' or ! &'( % ' rather than ! &'( % ! "#&'( as it is usually too complicated to be solved.

Note: not all solutions will be accepted. Check using the graphs.

3) Find ) given !) and ! "#, ! "#!) % )* +, % +-,.

4) Find ! given !) and )"#, !))"# % ! . Note that there is no way to find +- by this method unless the question gives +- .

5) Functions that involves

Important formula

0,

, 0

x x x

x x

!!= "

"

!= "

#

The equal sign in the inequality can be swapped at H2 level.

Example 1

( )f 2 , x x x= # $!

Sketching the graph

We can see that it is not one-one.

However, once we had restricted the domain of f , there will be no need to for the .

( )

2 , 2 2 ,2

2 , 2 0

2

2

0

2 ,

x x x x x

x x x x

! !# # #! !# = =" "

# # # < # + ( ) ( )2

2

, ,

g

0

f 2 2 x x

x

x

x x x

=

=

= # $

= # +

Note: 2

x x% unless the domain for is given!

$

'

.

.


16/27

Page 1 of 2

Graphing techniques

Some common basic graphs

Logarithmic ( )ln y x=

Asymptote : 0 x =

Exponential x y e=

Asymptote : 0 y =

Ellipse ( ) ( )2 2

0 0

2 21

x x y y

a b

! !+ =

Circle

(subset ofellipse)

( ) ( )2 2

0 0

2 21

x x y y

r r

! !+ =

or

( ) ( )2 2 2

0 0 x x y y r! + ! =

Hyperbola 2 2

2 21

x y

a b! =

2 2

2 21

y x

b a= !

As x " ±# ,2 2

2 2

y x

b a"

Hence asymptotes are y b

xa

= ± .

2 2

2 21

y x

b a! =

2 2

2 21

y x

b a= +

As x " ±# ,2 2

2 2

y x

b a"

Hence asymptotes are y b

x

a

= ± .

Note:

a) You can differentiate the two hyperbolas by finding x- y intercepts. Since one has only

x –intercepts, the other only has y-intercepts.

b) You may need to do transformations for basic graphs. Example: ( )ln 2 y x= ! is

obtained by translating ln y x= 2 units in the positive direction of x-axis.

!

"

#

!

"

#

$ %

&"'( !')

*

*

&"'( !')

!

"

! +$

%"

! +$

%"

,%

%

!

"

! +$

%"

! +$

%

"

,$

$


17/27

Page 2 of 2

Sketching rational function ( )( )

P

Q

x y

x=

( )( )

P

Q y =

Steps to sketch the curve

1. Horizontal/Oblique asymptotes

( )( )

P

Q y = is proper

( )( )

P

Q y

x= is improper, do long division to get

( ) ( )( )RFQ

y x= +

Horizontal Asymptotes

0= y ( )F y x= is the horizontal/oblique asymptotes

2. Vertical asymptotes, solve ( )Q 0 x = 3.

x- y intercepts

4. Differentiation (find stationary points),d

0d

y=

5. Valid/Empty y values regions (depend on what the questions asked)

a.

Form a quadratic equation in terms of x b. Since x $! when sketching curves, discriminant 0 D % c.

Solving the inequality will give you the valid y values

Note:

a) If you have the valid/empty y values, you need not do differentiation to find turning

points since the boundary of the region will be the y values of turning points.

Number of real roots for ( )P 0 x =

Given two graphs ( )f y x=

and ( )g y x=

. Equating them together( ) ( ) ( ) ( ) ( )f g f g 0 P 0 x x x x x= ! = ! =!

The number of real roots for the equation ( )P 0 x = is the number of intersections betweenthe two graphs.

Type 1: ( )f y x= and ( )g y x= are both given. Manipulate ( ) ( )f g x x= to form ( )P 0 x =

and conclude that since the graphs of ( )f y x= and ( )g y x= intersects at n points.

( )P 0 x = has n real roots.

Type 2: ( )g y x= is not given. Manipulate ( )P 0 x = to get ( ) ( )f g x x= . Then sketch( )g y x= .

Type 3: (Advance) ( )g y x= has unknown constant k , different values of k will givedifferent number of intersections.


18/27

Test for Population Mean based on

PopulationVarianceknown?

Distribution SampleSize

Test Statistic

(1

)

Yes

Normal Any

(2)

Any

n 50

By CLT,

(3

)

No

Any

n 50

By CLT,

(4)

Normal Small


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H2 Mathematics Cheat Sheet by Sean Lim