11 x1 t16 04 areas (2013)
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11 x1 t16 04 areas (2013) 11 x1 t16 04 areas (2013) Presentation Transcript

  • (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*