SlideShare a Scribd company logo
1 of 72
Download to read offline
(1) Area Below x axis
y

Areas
y = f(x)

x
Areas

(1) Area Below x axis
y

y = f(x)
A1

a

b

x
Areas

(1) Area Below x axis
y

y = f(x)
A1

a

 f  x dx  0
b

a

b

x
Areas

(1) Area Below x axis
y

y = f(x)
A1

a

 f  x dx  0
b

a

 A1   f  x dx
b

a

b

x
Areas

(1) Area Below x axis
y

y = f(x)
A1

a

 f  x dx  0
b

a

 A1   f  x dx
b

a

b

A2

c

x
Areas

(1) Area Below x axis
y

y = f(x)
A1

a



b

a

f  x dx  0

 A1   f  x dx
b

a

b

A2

c

x

 f  x dx  0
c

b
Areas

(1) Area Below x axis
y

y = f(x)
A1

a



b

a

f  x dx  0

 A1   f  x dx
b

a

b

A2

c

x

 f  x dx  0
c

b

 A2    f  x dx
c

b
Area below x axis is given by;
Area below x axis is given by;
A    f  x dx
c

b
Area below x axis is given by;
A    f  x dx
c

b

OR



 f  x dx
c

b
Area below x axis is given by;
A    f  x dx
c

b

OR



 f  x dx
c

b

OR
  f  x dx
b

c
e.g. (i)

y  x3

y

-1

1

x
e.g. (i)

y  x3

y

0

1

-1

1

x

1

A    x dx   x 3 dx
3

0
e.g. (i)

y  x3

y

0

1

A    x dx   x 3 dx
1

3

0

1 40 1 41
  x 1  x 0
4
4
-1

1

x
e.g. (i)

y  x3

y

0

1

-1

1

x

1

A    x dx   x 3 dx
3

0

1 40 1 41
  x 1  x 0
4
4
1
1
4
1
  0   1   4  0
4
4
1 1
 
4 4
1
 units 2
2




e.g. (i)

y  x3

y

0

1

-1

1

x

OR using symmetry of odd function

1

A    x dx   x 3 dx
3

0

1 40 1 41
  x 1  x 0
4
4
1
1
4
1
  0   1   4  0
4
4
1 1
 
4 4
1
 units 2
2




e.g. (i)

y  x3

y

0

1

-1

1

x

OR using symmetry of odd function
1

A  2  x 3 dx
0

1

A    x dx   x 3 dx
3

0

1 40 1 41
  x 1  x 0
4
4
1
1
4
1
  0   1   4  0
4
4
1 1
 
4 4
1
 units 2
2




e.g. (i)

y  x3

y

0

1

-1

1

x

OR using symmetry of odd function
1

A  2  x 3 dx
0

1 41
 x 0
2

1

A    x dx   x 3 dx
3

0

1 40 1 41
  x 1  x 0
4
4
1
1
4
1
  0   1   4  0
4
4
1 1
 
4 4
1
 units 2
2




e.g. (i)

y  x3

y

0

1

-1

1

x

OR using symmetry of odd function
1

A  2  x 3 dx
0

1 41
 x 0
2
1
  4  0
1
2
1
 units 2
2

1

A    x dx   x 3 dx
3

0

1 40 1 41
  x 1  x 0
4
4
1
1
4
1
  0   1   4  0
4
4
1 1
 
4 4
1
 units 2
2




(ii)

y

-2

-1

y  x x  1 x  2 

x
(ii)

y

-2

y  x x  1 x  2 

x

-1

A   x  3 x  2 x dx   x 3  3 x 2  2 x dx
1

2

3

2

0

1
(ii)

y  x x  1 x  2 

y

-2

x

-1

A   x  3 x  2 x dx   x 3  3 x 2  2 x dx
1

3

0

2

1

2

1

1

1 x 4  x3  x 2   1 x 4  x3  x 2 

 2  4
0
4



(ii)

y  x x  1 x  2 

y

-2

x

-1

A   x  3 x  2 x dx   x 3  3 x 2  2 x dx
1

3

0

2

1

2

1

1

1 x 4  x3  x 2   1 x 4  x3  x 2 

 2  4
0
4



1
4
3
2  1
4
3
2
 2  1   1   1     2    2    2    0


 4
4
1
 units 2
2
(2) Area On The y axis

y
y = f(x)
(b,d)
(a,c)

x
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)
(a,c)

x
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates

(a,c)

x
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates
d

3 A   g  y dy
c

(a,c)

x
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates
d

3 A   g  y dy
c

(a,c)
x
e.g.

y  x4

y

1

2

x
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates
d

3 A   g  y dy
c

(a,c)
x
e.g.

y  x4

y

x y

1

2

x

1
4
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates
d

3 A   g  y dy
c

(a,c)
16

A   y dy

x

1

e.g.

y  x4

y

x y

1

2

x

1
4

1
4
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates
d

3 A   g  y dy
c

(a,c)
16

A   y dy

x

1

e.g.

yx

y

x y

1

2

x

1
4

4
1
4

5 16
4

4 
 y 
5  1
(2) Area On The y axis

(1) Make x the subject
i.e. x = g(y)

y
y = f(x)
(b,d)

(2) Substitute the y coordinates
d

3 A   g  y dy
c

(a,c)
16

A   y dy

x

1

e.g.

yx

y

x y

1
4

4
1
4

5 16
4

4 
 y 
5  1

5
5
4 4 4
 16  1 
5


1

2

x

124
units 2

5
(3) Area Between Two Curves

y

x
(3) Area Between Two Curves

y

y = f(x)…(1)

x
(3) Area Between Two Curves

y

y = g(x)…(2)

y = f(x)…(1)

x
(3) Area Between Two Curves

y

y = g(x)…(2)

a

b

y = f(x)…(1)

x
(3) Area Between Two Curves

y

y = g(x)…(2)

a

b

y = f(x)…(1)

x

Area = Area under (1) – Area under (2)
(3) Area Between Two Curves

y

y = g(x)…(2)

a

b

y = f(x)…(1)

x

Area = Area under (1) – Area under (2)
b

b

a

a

  f  x dx   g  x dx
(3) Area Between Two Curves

y

y = g(x)…(2)

a

b

y = f(x)…(1)

x

Area = Area under (1) – Area under (2)
b

b

a
b

a

  f  x dx   g  x dx
   f  x   g  x dx
a
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
y

y  x5

yx

x
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
y

y  x5

x x
5

yx

x
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
y

y  x5

x x
x5  x  0
5

yx

x

xx 4  1  0
x  0 or x  1
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
y

y  x5

yx

A   x  x 5 dx
1

x x
x5  x  0
5

x

xx 4  1  0
x  0 or x  1

0
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
y

y  x5

yx

A   x  x 5 dx
1

x x
x5  x  0
5

x

xx 4  1  0
x  0 or x  1

0

1

1 2 1 6 
 x  x 
6 0
2
e.g. Find the area enclosed between the curves y  x 5 and y  x
in the positive quadrant.
y

y  x5

yx

A   x  x 5 dx
1

x x
x5  x  0
5

x

xx 4  1  0
x  0 or x  1

0

1

1 2 1 6 
 x  x 
6 0
2
1 1 2  1 1 6   0
  

6
2

1
 unit 2
3
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A

(2)
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A
To find points of intersection, solve simultaneously

(2)
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A
To find points of intersection, solve simultaneously
x  4  x2  4x

(2)
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A
To find points of intersection, solve simultaneously
x  4  x2  4x
x2  5x  4  0

(2)
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A
To find points of intersection, solve simultaneously
x  4  x2  4x
x2  5x  4  0
 x  4  x  1  0

(2)
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A
To find points of intersection, solve simultaneously
x  4  x2  4x
x2  5x  4  0
 x  4  x  1  0

x  1 or x  4

(2)
2002 HSC Question 4d)

The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
(i) Find the coordinates of A

To find points of intersection, solve simultaneously
x  4  x2  4x
x2  5x  4  0
 x  4  x  1  0
x  1 or x  4
 A is (1, 3)

(2)
(ii) Find the area of the shaded region bounded by y  x 2  4 x and

y  x  4.

(3)
(ii) Find the area of the shaded region bounded by y  x 2  4 x and

y  x  4.

4

A    x  4   x 2  4 x  dx
1

(3)
(ii) Find the area of the shaded region bounded by y  x 2  4 x and

y  x  4.

4

A    x  4   x 2  4 x  dx
1
4

    x 2  5 x  4 dx
1

(3)
(ii) Find the area of the shaded region bounded by y  x 2  4 x and

y  x  4.

4

A    x  4   x 2  4 x  dx
1
4

    x 2  5 x  4 dx
1

4

  1 x3  5 x 2  4 x 

1
2
 3


(3)
(ii) Find the area of the shaded region bounded by y  x 2  4 x and

y  x  4.

(3)

4

A    x  4   x 2  4 x  dx
1
4

    x 2  5 x  4 dx
1

4

  1 x3  5 x 2  4 x 

1
2
 3

1 3 5 2
1 3 5 2
   4    4   4  4    1  1  4 1
3
2
3
2




(ii) Find the area of the shaded region bounded by y  x 2  4 x and
y  x  4.
4
A    x  4   x 2  4 x  dx

(3)

1
4

    x 2  5 x  4 dx
1

4

  1 x3  5 x 2  4 x 

1
2
 3

1 3 5 2
1 3 5 2
   4    4   4  4    1  1  4 1
3
2
3
2





9
units 2
2


2005 HSC Question 8b)

(3)

The shaded region in the diagram is bounded by the circle of radius 2
centred at the origin, the parabola y  x 2  3 x  2 , and the x axis.
By considering the difference of two areas, find the area of the shaded
region.
2005 HSC Question 8b)

(3)

The shaded region in the diagram is bounded by the circle of radius 2
centred at the origin, the parabola y  x 2  3 x  2 , and the x axis.
By considering the difference of two areas, find the area of the shaded
region.
Note: area must be broken up into two areas, due to the different
boundaries.
2005 HSC Question 8b)

(3)

The shaded region in the diagram is bounded by the circle of radius 2
centred at the origin, the parabola y  x 2  3 x  2 , and the x axis.
By considering the difference of two areas, find the area of the shaded
region.
Note: area must be broken up into two areas, due to the different
boundaries.
2005 HSC Question 8b)

(3)

The shaded region in the diagram is bounded by the circle of radius 2
centred at the origin, the parabola y  x 2  3 x  2 , and the x axis.
By considering the difference of two areas, find the area of the shaded
region.
Note: area must be broken up into two areas, due to the different
boundaries.
Area between circle and parabola
2005 HSC Question 8b)

(3)

The shaded region in the diagram is bounded by the circle of radius 2
centred at the origin, the parabola y  x 2  3 x  2 , and the x axis.
By considering the difference of two areas, find the area of the shaded
region.
Note: area must be broken up into two areas, due to the different
boundaries.
Area between circle and parabola and area between circle and x axis
It is easier to subtract the area under the parabola from the quadrant.
It is easier to subtract the area under the parabola from the quadrant.
1
1
2
A    2     x 2  3 x  2 dx
4
0
It is easier to subtract the area under the parabola from the quadrant.
1
1
2
A    2     x 2  3 x  2 dx
4
0
1
 1 x3  3 x 2  2 x 
 
0
2
3

It is easier to subtract the area under the parabola from the quadrant.
1
1
2
A    2     x 2  3 x  2 dx
4
0
1
 1 x3  3 x 2  2 x 
 
0
2
3

1 3 3 2
   1  1  2 1  0
3
2




It is easier to subtract the area under the parabola from the quadrant.
1
1
2
A    2     x 2  3 x  2 dx
4
0
1
 1 x3  3 x 2  2 x 
 
0
2
3

1 3 3 2
   1  1  2 1  0
3
2
   5  units 2


6





Exercise 11E; 2bceh, 3bd, 4bd, 5bd, 7begj, 8d, 9a, 11, 18*

Exercise 11F; 1bdeh, 4bd, 7d, 10, 11b, 13, 15*

More Related Content

What's hot

CAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONS
CAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONSCAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONS
CAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONS
Carlon Baird
 
X2 t01 03 argand diagram (2013)
X2 t01 03 argand diagram (2013)X2 t01 03 argand diagram (2013)
X2 t01 03 argand diagram (2013)
Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
Nigel Simmons
 
Modul bimbingan add maths
Modul bimbingan add mathsModul bimbingan add maths
Modul bimbingan add maths
Sasi Villa
 
F4 02 Quadratic Expressions And
F4 02 Quadratic Expressions AndF4 02 Quadratic Expressions And
F4 02 Quadratic Expressions And
guestcc333c
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
Nigel Simmons
 
Pc12 sol c04_4-3
Pc12 sol c04_4-3Pc12 sol c04_4-3
Pc12 sol c04_4-3
Garden City
 
11 X1 T01 08 Simultaneous Equations (2010)
11 X1 T01 08 Simultaneous Equations (2010)11 X1 T01 08 Simultaneous Equations (2010)
11 X1 T01 08 Simultaneous Equations (2010)
Nigel Simmons
 

What's hot (20)

CAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONS
CAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONSCAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONS
CAPE PURE MATHEMATICS UNIT 2 MODULE 2 PRACTICE QUESTIONS
 
Pure Mathematics Unit 2 - Textbook
Pure Mathematics Unit 2 - TextbookPure Mathematics Unit 2 - Textbook
Pure Mathematics Unit 2 - Textbook
 
X2 t01 03 argand diagram (2013)
X2 t01 03 argand diagram (2013)X2 t01 03 argand diagram (2013)
X2 t01 03 argand diagram (2013)
 
Unit 7 Review A
Unit 7 Review AUnit 7 Review A
Unit 7 Review A
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
Ejercicios resueltos
Ejercicios resueltosEjercicios resueltos
Ejercicios resueltos
 
Tricky log graphs
Tricky log graphsTricky log graphs
Tricky log graphs
 
0207 ch 2 day 7
0207 ch 2 day 70207 ch 2 day 7
0207 ch 2 day 7
 
Function graphs
Function graphsFunction graphs
Function graphs
 
Succesive differntiation
Succesive differntiationSuccesive differntiation
Succesive differntiation
 
Cbse Class 12 Maths Sample Paper 2012
Cbse Class 12 Maths Sample Paper 2012Cbse Class 12 Maths Sample Paper 2012
Cbse Class 12 Maths Sample Paper 2012
 
Multiple Choice Questions_Successive Differentiation (CALCULUS)
Multiple Choice Questions_Successive Differentiation (CALCULUS)Multiple Choice Questions_Successive Differentiation (CALCULUS)
Multiple Choice Questions_Successive Differentiation (CALCULUS)
 
Modul bimbingan add maths
Modul bimbingan add mathsModul bimbingan add maths
Modul bimbingan add maths
 
Orthogonal Range Searching
Orthogonal Range SearchingOrthogonal Range Searching
Orthogonal Range Searching
 
F4 02 Quadratic Expressions And
F4 02 Quadratic Expressions AndF4 02 Quadratic Expressions And
F4 02 Quadratic Expressions And
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
Class 12 practice paper
Class 12 practice paperClass 12 practice paper
Class 12 practice paper
 
Pc12 sol c04_4-3
Pc12 sol c04_4-3Pc12 sol c04_4-3
Pc12 sol c04_4-3
 
Modul 1 functions
Modul 1 functionsModul 1 functions
Modul 1 functions
 
11 X1 T01 08 Simultaneous Equations (2010)
11 X1 T01 08 Simultaneous Equations (2010)11 X1 T01 08 Simultaneous Equations (2010)
11 X1 T01 08 Simultaneous Equations (2010)
 

Similar to 11 x1 t16 04 areas (2013)

11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
Nigel Simmons
 
Area and volume practice
Area and volume practiceArea and volume practice
Area and volume practice
tschmucker
 
11 x1 t16 01 area under curve (2013)
11 x1 t16 01 area under curve (2013)11 x1 t16 01 area under curve (2013)
11 x1 t16 01 area under curve (2013)
Nigel Simmons
 
X2 T06 02 Cylindrical Shells
X2 T06 02 Cylindrical ShellsX2 T06 02 Cylindrical Shells
X2 T06 02 Cylindrical Shells
Nigel Simmons
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
Nigel Simmons
 
Pc12 sol c04_review
Pc12 sol c04_reviewPc12 sol c04_review
Pc12 sol c04_review
Garden City
 
Pc12 sol c04_4-1
Pc12 sol c04_4-1Pc12 sol c04_4-1
Pc12 sol c04_4-1
Garden City
 

Similar to 11 x1 t16 04 areas (2013) (20)

11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
Area and volume practice
Area and volume practiceArea and volume practice
Area and volume practice
 
Grph quad fncts
Grph quad fnctsGrph quad fncts
Grph quad fncts
 
Higher formal homeworks unit 1
Higher formal homeworks   unit 1Higher formal homeworks   unit 1
Higher formal homeworks unit 1
 
11 x1 t16 01 area under curve (2013)
11 x1 t16 01 area under curve (2013)11 x1 t16 01 area under curve (2013)
11 x1 t16 01 area under curve (2013)
 
X2 T06 02 Cylindrical Shells
X2 T06 02 Cylindrical ShellsX2 T06 02 Cylindrical Shells
X2 T06 02 Cylindrical Shells
 
Ejercicios radhames ultima unidad
Ejercicios radhames ultima unidadEjercicios radhames ultima unidad
Ejercicios radhames ultima unidad
 
Calc 7.2b
Calc 7.2bCalc 7.2b
Calc 7.2b
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
 
Pc12 sol c04_review
Pc12 sol c04_reviewPc12 sol c04_review
Pc12 sol c04_review
 
Review
ReviewReview
Review
 
Form 4 add maths note
Form 4 add maths noteForm 4 add maths note
Form 4 add maths note
 
El6303 solu 3 f15 1
El6303 solu 3 f15  1 El6303 solu 3 f15  1
El6303 solu 3 f15 1
 
Ejercicios victor
Ejercicios victorEjercicios victor
Ejercicios victor
 
Pc12 sol c04_4-1
Pc12 sol c04_4-1Pc12 sol c04_4-1
Pc12 sol c04_4-1
 
Module 3 quadratic functions
Module 3   quadratic functionsModule 3   quadratic functions
Module 3 quadratic functions
 
1603 plane sections of real and complex tori
1603 plane sections of real and complex tori1603 plane sections of real and complex tori
1603 plane sections of real and complex tori
 
Assignment For Matlab Report Subject Calculus 2
Assignment For Matlab Report Subject  Calculus 2Assignment For Matlab Report Subject  Calculus 2
Assignment For Matlab Report Subject Calculus 2
 
Μαθηματικά Γ Λυκείου - Ν. Ράπτης
Μαθηματικά Γ Λυκείου - Ν. ΡάπτηςΜαθηματικά Γ Λυκείου - Ν. Ράπτης
Μαθηματικά Γ Λυκείου - Ν. Ράπτης
 
Areas under curve
Areas under curveAreas under curve
Areas under curve
 

More from Nigel Simmons

12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
Nigel Simmons
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
Nigel Simmons
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
Nigel Simmons
 
X2 t01 11 nth roots of unity (2012)
X2 t01 11 nth roots of unity (2012)X2 t01 11 nth roots of unity (2012)
X2 t01 11 nth roots of unity (2012)
Nigel Simmons
 
X2 t01 10 complex & trig (2013)
X2 t01 10 complex & trig (2013)X2 t01 10 complex & trig (2013)
X2 t01 10 complex & trig (2013)
Nigel Simmons
 
X2 t01 09 de moivres theorem
X2 t01 09 de moivres theoremX2 t01 09 de moivres theorem
X2 t01 09 de moivres theorem
Nigel Simmons
 
X2 t01 08 locus & complex nos 2 (2013)
X2 t01 08  locus & complex nos 2 (2013)X2 t01 08  locus & complex nos 2 (2013)
X2 t01 08 locus & complex nos 2 (2013)
Nigel Simmons
 
X2 t01 07 locus & complex nos 1 (2013)
X2 t01 07 locus & complex nos 1 (2013)X2 t01 07 locus & complex nos 1 (2013)
X2 t01 07 locus & complex nos 1 (2013)
Nigel Simmons
 
X2 t01 06 geometrical representation (2013)
X2 t01 06 geometrical representation (2013)X2 t01 06 geometrical representation (2013)
X2 t01 06 geometrical representation (2013)
Nigel Simmons
 

More from Nigel Simmons (20)

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
 
X2 t01 11 nth roots of unity (2012)
X2 t01 11 nth roots of unity (2012)X2 t01 11 nth roots of unity (2012)
X2 t01 11 nth roots of unity (2012)
 
X2 t01 10 complex & trig (2013)
X2 t01 10 complex & trig (2013)X2 t01 10 complex & trig (2013)
X2 t01 10 complex & trig (2013)
 
X2 t01 09 de moivres theorem
X2 t01 09 de moivres theoremX2 t01 09 de moivres theorem
X2 t01 09 de moivres theorem
 
X2 t01 08 locus & complex nos 2 (2013)
X2 t01 08  locus & complex nos 2 (2013)X2 t01 08  locus & complex nos 2 (2013)
X2 t01 08 locus & complex nos 2 (2013)
 
X2 t01 07 locus & complex nos 1 (2013)
X2 t01 07 locus & complex nos 1 (2013)X2 t01 07 locus & complex nos 1 (2013)
X2 t01 07 locus & complex nos 1 (2013)
 
X2 t01 06 geometrical representation (2013)
X2 t01 06 geometrical representation (2013)X2 t01 06 geometrical representation (2013)
X2 t01 06 geometrical representation (2013)
 

Recently uploaded

Salient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functionsSalient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functions
KarakKing
 

Recently uploaded (20)

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...
Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...
Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...
 
Sociology 101 Demonstration of Learning Exhibit
Sociology 101 Demonstration of Learning ExhibitSociology 101 Demonstration of Learning Exhibit
Sociology 101 Demonstration of Learning Exhibit
 
How to Give a Domain for a Field in Odoo 17
How to Give a Domain for a Field in Odoo 17How to Give a Domain for a Field in Odoo 17
How to Give a Domain for a Field in Odoo 17
 
Jamworks pilot and AI at Jisc (20/03/2024)
Jamworks pilot and AI at Jisc (20/03/2024)Jamworks pilot and AI at Jisc (20/03/2024)
Jamworks pilot and AI at Jisc (20/03/2024)
 
Salient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functionsSalient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functions
 
Plant propagation: Sexual and Asexual propapagation.pptx
Plant propagation: Sexual and Asexual propapagation.pptxPlant propagation: Sexual and Asexual propapagation.pptx
Plant propagation: Sexual and Asexual propapagation.pptx
 
Kodo Millet PPT made by Ghanshyam bairwa college of Agriculture kumher bhara...
Kodo Millet  PPT made by Ghanshyam bairwa college of Agriculture kumher bhara...Kodo Millet  PPT made by Ghanshyam bairwa college of Agriculture kumher bhara...
Kodo Millet PPT made by Ghanshyam bairwa college of Agriculture kumher bhara...
 
80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...
80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...
80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...
 
Single or Multiple melodic lines structure
Single or Multiple melodic lines structureSingle or Multiple melodic lines structure
Single or Multiple melodic lines structure
 
Understanding Accommodations and Modifications
Understanding  Accommodations and ModificationsUnderstanding  Accommodations and Modifications
Understanding Accommodations and Modifications
 
On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx
On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptxOn_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx
On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx
 
Mehran University Newsletter Vol-X, Issue-I, 2024
Mehran University Newsletter Vol-X, Issue-I, 2024Mehran University Newsletter Vol-X, Issue-I, 2024
Mehran University Newsletter Vol-X, Issue-I, 2024
 
Basic Civil Engineering first year Notes- Chapter 4 Building.pptx
Basic Civil Engineering first year Notes- Chapter 4 Building.pptxBasic Civil Engineering first year Notes- Chapter 4 Building.pptx
Basic Civil Engineering first year Notes- Chapter 4 Building.pptx
 
ICT role in 21st century education and it's challenges.
ICT role in 21st century education and it's challenges.ICT role in 21st century education and it's challenges.
ICT role in 21st century education and it's challenges.
 
Food safety_Challenges food safety laboratories_.pdf
Food safety_Challenges food safety laboratories_.pdfFood safety_Challenges food safety laboratories_.pdf
Food safety_Challenges food safety laboratories_.pdf
 
SOC 101 Demonstration of Learning Presentation
SOC 101 Demonstration of Learning PresentationSOC 101 Demonstration of Learning Presentation
SOC 101 Demonstration of Learning Presentation
 
Interdisciplinary_Insights_Data_Collection_Methods.pptx
Interdisciplinary_Insights_Data_Collection_Methods.pptxInterdisciplinary_Insights_Data_Collection_Methods.pptx
Interdisciplinary_Insights_Data_Collection_Methods.pptx
 
Towards a code of practice for AI in AT.pptx
Towards a code of practice for AI in AT.pptxTowards a code of practice for AI in AT.pptx
Towards a code of practice for AI in AT.pptx
 
NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...
NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...
NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...
 
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptxHMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
 

11 x1 t16 04 areas (2013)

  • 1. (1) Area Below x axis y Areas y = f(x) x
  • 2. Areas (1) Area Below x axis y y = f(x) A1 a b x
  • 3. Areas (1) Area Below x axis y y = f(x) A1 a  f  x dx  0 b a b x
  • 4. Areas (1) Area Below x axis y y = f(x) A1 a  f  x dx  0 b a  A1   f  x dx b a b x
  • 5. Areas (1) Area Below x axis y y = f(x) A1 a  f  x dx  0 b a  A1   f  x dx b a b A2 c x
  • 6. Areas (1) Area Below x axis y y = f(x) A1 a  b a f  x dx  0  A1   f  x dx b a b A2 c x  f  x dx  0 c b
  • 7. Areas (1) Area Below x axis y y = f(x) A1 a  b a f  x dx  0  A1   f  x dx b a b A2 c x  f  x dx  0 c b  A2    f  x dx c b
  • 8. Area below x axis is given by;
  • 9. Area below x axis is given by; A    f  x dx c b
  • 10. Area below x axis is given by; A    f  x dx c b OR   f  x dx c b
  • 11. Area below x axis is given by; A    f  x dx c b OR   f  x dx c b OR   f  x dx b c
  • 12. e.g. (i) y  x3 y -1 1 x
  • 13. e.g. (i) y  x3 y 0 1 -1 1 x 1 A    x dx   x 3 dx 3 0
  • 14. e.g. (i) y  x3 y 0 1 A    x dx   x 3 dx 1 3 0 1 40 1 41   x 1  x 0 4 4 -1 1 x
  • 15. e.g. (i) y  x3 y 0 1 -1 1 x 1 A    x dx   x 3 dx 3 0 1 40 1 41   x 1  x 0 4 4 1 1 4 1   0   1   4  0 4 4 1 1   4 4 1  units 2 2  
  • 16. e.g. (i) y  x3 y 0 1 -1 1 x OR using symmetry of odd function 1 A    x dx   x 3 dx 3 0 1 40 1 41   x 1  x 0 4 4 1 1 4 1   0   1   4  0 4 4 1 1   4 4 1  units 2 2  
  • 17. e.g. (i) y  x3 y 0 1 -1 1 x OR using symmetry of odd function 1 A  2  x 3 dx 0 1 A    x dx   x 3 dx 3 0 1 40 1 41   x 1  x 0 4 4 1 1 4 1   0   1   4  0 4 4 1 1   4 4 1  units 2 2  
  • 18. e.g. (i) y  x3 y 0 1 -1 1 x OR using symmetry of odd function 1 A  2  x 3 dx 0 1 41  x 0 2 1 A    x dx   x 3 dx 3 0 1 40 1 41   x 1  x 0 4 4 1 1 4 1   0   1   4  0 4 4 1 1   4 4 1  units 2 2  
  • 19. e.g. (i) y  x3 y 0 1 -1 1 x OR using symmetry of odd function 1 A  2  x 3 dx 0 1 41  x 0 2 1   4  0 1 2 1  units 2 2 1 A    x dx   x 3 dx 3 0 1 40 1 41   x 1  x 0 4 4 1 1 4 1   0   1   4  0 4 4 1 1   4 4 1  units 2 2  
  • 20. (ii) y -2 -1 y  x x  1 x  2  x
  • 21. (ii) y -2 y  x x  1 x  2  x -1 A   x  3 x  2 x dx   x 3  3 x 2  2 x dx 1 2 3 2 0 1
  • 22. (ii) y  x x  1 x  2  y -2 x -1 A   x  3 x  2 x dx   x 3  3 x 2  2 x dx 1 3 0 2 1 2 1 1 1 x 4  x3  x 2   1 x 4  x3  x 2    2  4 0 4   
  • 23. (ii) y  x x  1 x  2  y -2 x -1 A   x  3 x  2 x dx   x 3  3 x 2  2 x dx 1 3 0 2 1 2 1 1 1 x 4  x3  x 2   1 x 4  x3  x 2    2  4 0 4    1 4 3 2  1 4 3 2  2  1   1   1     2    2    2    0    4 4 1  units 2 2
  • 24. (2) Area On The y axis y y = f(x) (b,d) (a,c) x
  • 25. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (a,c) x
  • 26. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates (a,c) x
  • 27. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates d 3 A   g  y dy c (a,c) x
  • 28. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates d 3 A   g  y dy c (a,c) x e.g. y  x4 y 1 2 x
  • 29. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates d 3 A   g  y dy c (a,c) x e.g. y  x4 y x y 1 2 x 1 4
  • 30. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates d 3 A   g  y dy c (a,c) 16 A   y dy x 1 e.g. y  x4 y x y 1 2 x 1 4 1 4
  • 31. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates d 3 A   g  y dy c (a,c) 16 A   y dy x 1 e.g. yx y x y 1 2 x 1 4 4 1 4 5 16 4 4   y  5  1
  • 32. (2) Area On The y axis (1) Make x the subject i.e. x = g(y) y y = f(x) (b,d) (2) Substitute the y coordinates d 3 A   g  y dy c (a,c) 16 A   y dy x 1 e.g. yx y x y 1 4 4 1 4 5 16 4 4   y  5  1 5 5 4 4 4  16  1  5  1 2 x 124 units 2  5
  • 33. (3) Area Between Two Curves y x
  • 34. (3) Area Between Two Curves y y = f(x)…(1) x
  • 35. (3) Area Between Two Curves y y = g(x)…(2) y = f(x)…(1) x
  • 36. (3) Area Between Two Curves y y = g(x)…(2) a b y = f(x)…(1) x
  • 37. (3) Area Between Two Curves y y = g(x)…(2) a b y = f(x)…(1) x Area = Area under (1) – Area under (2)
  • 38. (3) Area Between Two Curves y y = g(x)…(2) a b y = f(x)…(1) x Area = Area under (1) – Area under (2) b b a a   f  x dx   g  x dx
  • 39. (3) Area Between Two Curves y y = g(x)…(2) a b y = f(x)…(1) x Area = Area under (1) – Area under (2) b b a b a   f  x dx   g  x dx    f  x   g  x dx a
  • 40. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant.
  • 41. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant. y y  x5 yx x
  • 42. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant. y y  x5 x x 5 yx x
  • 43. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant. y y  x5 x x x5  x  0 5 yx x xx 4  1  0 x  0 or x  1
  • 44. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant. y y  x5 yx A   x  x 5 dx 1 x x x5  x  0 5 x xx 4  1  0 x  0 or x  1 0
  • 45. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant. y y  x5 yx A   x  x 5 dx 1 x x x5  x  0 5 x xx 4  1  0 x  0 or x  1 0 1 1 2 1 6   x  x  6 0 2
  • 46. e.g. Find the area enclosed between the curves y  x 5 and y  x in the positive quadrant. y y  x5 yx A   x  x 5 dx 1 x x x5  x  0 5 x xx 4  1  0 x  0 or x  1 0 1 1 2 1 6   x  x  6 0 2 1 1 2  1 1 6   0     6 2  1  unit 2 3
  • 47. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A.
  • 48. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A (2)
  • 49. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A To find points of intersection, solve simultaneously (2)
  • 50. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A To find points of intersection, solve simultaneously x  4  x2  4x (2)
  • 51. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A To find points of intersection, solve simultaneously x  4  x2  4x x2  5x  4  0 (2)
  • 52. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A To find points of intersection, solve simultaneously x  4  x2  4x x2  5x  4  0  x  4  x  1  0 (2)
  • 53. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A To find points of intersection, solve simultaneously x  4  x2  4x x2  5x  4  0  x  4  x  1  0 x  1 or x  4 (2)
  • 54. 2002 HSC Question 4d) The graphs of y  x  4 and y  x 2  4 x intersect at the points  4,0  A. (i) Find the coordinates of A To find points of intersection, solve simultaneously x  4  x2  4x x2  5x  4  0  x  4  x  1  0 x  1 or x  4  A is (1, 3) (2)
  • 55. (ii) Find the area of the shaded region bounded by y  x 2  4 x and y  x  4. (3)
  • 56. (ii) Find the area of the shaded region bounded by y  x 2  4 x and y  x  4. 4 A    x  4   x 2  4 x  dx 1 (3)
  • 57. (ii) Find the area of the shaded region bounded by y  x 2  4 x and y  x  4. 4 A    x  4   x 2  4 x  dx 1 4     x 2  5 x  4 dx 1 (3)
  • 58. (ii) Find the area of the shaded region bounded by y  x 2  4 x and y  x  4. 4 A    x  4   x 2  4 x  dx 1 4     x 2  5 x  4 dx 1 4   1 x3  5 x 2  4 x   1 2  3  (3)
  • 59. (ii) Find the area of the shaded region bounded by y  x 2  4 x and y  x  4. (3) 4 A    x  4   x 2  4 x  dx 1 4     x 2  5 x  4 dx 1 4   1 x3  5 x 2  4 x   1 2  3  1 3 5 2 1 3 5 2    4    4   4  4    1  1  4 1 3 2 3 2  
  • 60. (ii) Find the area of the shaded region bounded by y  x 2  4 x and y  x  4. 4 A    x  4   x 2  4 x  dx (3) 1 4     x 2  5 x  4 dx 1 4   1 x3  5 x 2  4 x   1 2  3  1 3 5 2 1 3 5 2    4    4   4  4    1  1  4 1 3 2 3 2   9 units 2 2 
  • 61. 2005 HSC Question 8b) (3) The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola y  x 2  3 x  2 , and the x axis. By considering the difference of two areas, find the area of the shaded region.
  • 62. 2005 HSC Question 8b) (3) The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola y  x 2  3 x  2 , and the x axis. By considering the difference of two areas, find the area of the shaded region. Note: area must be broken up into two areas, due to the different boundaries.
  • 63. 2005 HSC Question 8b) (3) The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola y  x 2  3 x  2 , and the x axis. By considering the difference of two areas, find the area of the shaded region. Note: area must be broken up into two areas, due to the different boundaries.
  • 64. 2005 HSC Question 8b) (3) The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola y  x 2  3 x  2 , and the x axis. By considering the difference of two areas, find the area of the shaded region. Note: area must be broken up into two areas, due to the different boundaries. Area between circle and parabola
  • 65. 2005 HSC Question 8b) (3) The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola y  x 2  3 x  2 , and the x axis. By considering the difference of two areas, find the area of the shaded region. Note: area must be broken up into two areas, due to the different boundaries. Area between circle and parabola and area between circle and x axis
  • 66.
  • 67. It is easier to subtract the area under the parabola from the quadrant.
  • 68. It is easier to subtract the area under the parabola from the quadrant. 1 1 2 A    2     x 2  3 x  2 dx 4 0
  • 69. It is easier to subtract the area under the parabola from the quadrant. 1 1 2 A    2     x 2  3 x  2 dx 4 0 1  1 x3  3 x 2  2 x    0 2 3 
  • 70. It is easier to subtract the area under the parabola from the quadrant. 1 1 2 A    2     x 2  3 x  2 dx 4 0 1  1 x3  3 x 2  2 x    0 2 3  1 3 3 2    1  1  2 1  0 3 2  
  • 71. It is easier to subtract the area under the parabola from the quadrant. 1 1 2 A    2     x 2  3 x  2 dx 4 0 1  1 x3  3 x 2  2 x    0 2 3  1 3 3 2    1  1  2 1  0 3 2    5  units 2   6   
  • 72. Exercise 11E; 2bceh, 3bd, 4bd, 5bd, 7begj, 8d, 9a, 11, 18* Exercise 11F; 1bdeh, 4bd, 7d, 10, 11b, 13, 15*