The document discusses two methods for calculating the area between a curve and the x- or y-axis. 1) For the area below the x-axis (A1), it is given by the integral of the function f(x) between the bounds a and b. 2) For the area on the y-axis between coordinates (a,c) and (b,d), it involves making x the subject of the equation (x=g(y)), then calculating the integral of g(y) between c and d. Examples are given to demonstrate these methods.

1. (1) Area Below x axis
Areas
y
y = f(x)
x

2. (1) Area Below x axis
Areas
y
y = f(x)
A1
a b x

3. (1) Area Below x axis
Areas
y
y = f(x)
A1
a b x
 f  x dx  0
b
a

4. (1) Area Below x axis
Areas
y
y = f(x)
A1
a b x
 f  x dx  0
b
a
 A1   f  x dx
b
a

5. (1) Area Below x axis
Areas
y
y = f(x)
A1
c x
a b A2
 f  x dx  0
b
a
 A1   f  x dx
b
a

6. (1) Area Below x axis
Areas
y
y = f(x)
A1
c x
a b A2
 f  x dx  0
c
f  x dx  0
b
a b
 A1   f  x dx
b
a

7. (1) Area Below x axis
Areas
y
y = f(x)
A1
c x
a b A2
 f  x dx  0
c
f  x dx  0
b
a b
 A2    f  x dx
c
 A1   f  x dx
b
a 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 y  x3
-1 1 x

13. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0
-1 1 x

14. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0

4
x 1  4 x 0
1 40 1 41
-1 1 x

15. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0

4
x 1  4 x 0
1 40 1 41
-1 1 x 1
4
 4

  0   1   4  0
1
4
1
1 1
 
4 4
1
 units 2
2

16. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0

4
x 1  4 x 0
1 40 1 41
-1 1 x 1
4
 4

  0   1   4  0
1
4
1
1 1
 
OR using symmetry of odd function 4 4
1
 units 2
2

17. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0

4
x 1  4 x 0
1 40 1 41
-1 1 x 1
4
 4

  0   1   4  0
1
4
1
1 1
 
OR using symmetry of odd function 4 4
1
1
A  2  x 3 dx  units 2
0 2

18. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0

4
x 1  4 x 0
1 40 1 41
-1 1 x 1
4
 4

  0   1   4  0
1
4
1
1 1
 
OR using symmetry of odd function 4 4
1
1
A  2  x 3 dx  units 2
0 2
 x 0
1 41
2

19. e.g. (i) y y  x3 0 1
A    x dx   x 3 dx
3
1 0

4
x 1  4 x 0
1 40 1 41
-1 1 x 1
4
 4

  0   1   4  0
1
4
1
1 1
 
OR using symmetry of odd function 4 4
1
1
A  2  x 3 dx  units 2
0 2
 x 0
1 41
2
  4  0
1
1
2
1
 units 2
2

20. (ii) y y  x x  1 x  2 
-2 -1 x

21. (ii) y y  x x  1 x  2 
-2 -1 x
A   x  3 x  2 x dx   x 3  3 x 2  2 x dx
1 0
3 2
2 1

22. (ii) y y  x x  1 x  2 
-2 -1 x
A   x  3 x  2 x dx   x 3  3 x 2  2 x dx
1 0
3 2
2 1
1 1
  x 4  x3  x 2    x 4  x3  x 2 
1 1
4
  2  4
  0


23. (ii) y y  x x  1 x  2 
-2 -1 x
A   x  3 x  2 x dx   x 3  3 x 2  2 x dx
1 0
3 2
2 1
1 1
  x 4  x3  x 2    x 4  x3  x 2 
1 1
4
  2  4
  0

 2 1  14   13   12    1  2 4   2 3   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
y i.e. x = g(y)
y = f(x)
(b,d)
(a,c)
x

26. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d)
(a,c)
x

27. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d) d
3 A   g  y dy
c
(a,c)
x

28. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d) d
3 A   g  y dy
c
(a,c)
x
e.g. y y  x4
1 2 x

29. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d) d
3 A   g  y dy
c
(a,c)
x
e.g. y y  x4
1
x y 4
1 2 x

30. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d) d
3 A   g  y dy
c
(a,c)
16 1
x A   y dy
4
1
e.g. y y  x4
1
x y 4
1 2 x

31. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d) d
3 A   g  y dy
c
(a,c)
16 1
x A   y dy
4
1
5 16
e.g. y yx 4
4 
1  y  4
5  1
x y 4
1 2 x

32. (2) Area On The y axis (1) Make x the subject
y i.e. x = g(y)
y = f(x) (2) Substitute the y coordinates
(b,d) d
3 A   g  y dy
c
(a,c)
16 1
x A   y dy
4
1
5 16
e.g. y yx 4
4 
1  y  4
5  1
x y 4
4  5 5
 16 4  14 
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) y = f(x)…(1)
a b x

37. (3) Area Between Two Curves
y y = g(x)…(2) y = f(x)…(1)
a b x
Area = Area under (1) – Area under (2)

38. (3) Area Between Two Curves
y y = g(x)…(2) y = f(x)…(1)
a b x
Area = Area under (1) – Area under (2)
b b
  f  x dx   g  x dx
a a

39. (3) Area Between Two Curves
y y = g(x)…(2) y = f(x)…(1)
a b x
Area = Area under (1) – Area under (2)
b b
  f  x dx   g  x dx
a a
b
   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 yx
x
x x
5

43. 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
x x
5
x5  x  0
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 0
x x
5
x5  x  0
xx 4  1  0
x  0 or x  1

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 0
x x
5
1
1 1 
x5  x  0   x2  x6 
2 6 0
xx 4  1  0
x  0 or x  1

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 0
x x
5
1
1 1 
x5  x  0   x2  x6 
2 6 0
xx 4  1  0
1 1 2  1 1 6   0
  
x  0 or x  1 
2 6 
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 (2)
To find points of intersection, solve simultaneously

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 (2)
To find points of intersection, solve simultaneously
x  4  x2  4x

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 (2)
To find points of intersection, solve simultaneously
x  4  x2  4x
x2  5x  4  0

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 (2)
To find points of intersection, solve simultaneously
x  4  x2  4x
x2  5x  4  0
 x  4  x  1  0

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 (2)
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

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 (2)
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)

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

56. (ii) Find the area of the shaded region bounded by y  x 2  4 x and (3)
y  x  4. 4
A    x  4   x 2  4 x  dx
1

57. (ii) Find the area of the shaded region bounded by y  x 2  4 x and (3)
y  x  4. 4
A    x  4   x 2  4 x  dx
1
4
    x 2  5 x  4 dx
1

58. (ii) Find the area of the shaded region bounded by y  x 2  4 x and (3)
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 

 3 2 1


59. (ii) Find the area of the shaded region bounded by y  x 2  4 x and (3)
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 
 3 2 1

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 (3)
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 
 3 2 1

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

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
A    2     x 2  3 x  2 dx
1 2
4 0

69. It is easier to subtract the area under the parabola from the quadrant.
1
A    2     x 2  3 x  2 dx
1 2
4 0
1
    x  x  2x
1 3 3 2
3
 2 0


70. It is easier to subtract the area under the parabola from the quadrant.
1
A    2     x 2  3 x  2 dx
1 2
4 0
1
    x  x  2x
1 3 3 2
3
 2 0

 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
A    2     x 2  3 x  2 dx
1 2
4 0
1
    x  x  2x
1 3 3 2
3
 2 0

 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*