SlideShare une entreprise Scribd logo
1  sur  72
Télécharger pour lire hors ligne
(1) Area Below x axis
                        Areas
           y
                                    y = f(x)




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


                        A1


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


                        A1


                    a        b       x


 f  x dx  0
 b

a
(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
(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
(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
(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
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       y  x3




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




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




       -2   -1         x
(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
(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
                                                 
(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
(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
 y                                  i.e. x = g(y)
     y = f(x)
           (b,d)


                   (a,c)

                           x
(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
(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
(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
(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
(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
(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
(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
(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)       y = f(x)…(1)




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




      a           b                      x

     Area = Area under (1) – Area under (2)
(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
(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
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    yx




                         x
             x x
              5
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
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
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
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
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                                               (2)
  To find points of intersection, solve simultaneously
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
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
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
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
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)
(ii) Find the area of the shaded region bounded by y  x 2  4 x and   (3)
   y  x  4.
(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
(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
(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
                                            
(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             
(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
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
                    A    2     x 2  3 x  2 dx
                         1      2

                         4         0
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
                                                
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            
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
Exercise 11E; 2bceh, 3bd, 4bd, 5bd, 7begj, 8d, 9a, 11, 18*


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

Contenu connexe

Tendances

X2 T07 02 transformations (2011)
X2 T07 02 transformations (2011)X2 T07 02 transformations (2011)
X2 T07 02 transformations (2011)Nigel Simmons
 
X2 T04 03 cuve sketching - addition, subtraction, multiplication and division
X2 T04 03 cuve sketching - addition, subtraction,  multiplication and divisionX2 T04 03 cuve sketching - addition, subtraction,  multiplication and division
X2 T04 03 cuve sketching - addition, subtraction, multiplication and divisionNigel Simmons
 
Formula List Math 1230
Formula List Math 1230Formula List Math 1230
Formula List Math 1230samhui48
 
X2 T04 04 curve sketching - reciprocal functions
X2 T04 04 curve sketching - reciprocal functionsX2 T04 04 curve sketching - reciprocal functions
X2 T04 04 curve sketching - reciprocal functionsNigel Simmons
 

Tendances (9)

Exercise #10 notes
Exercise #10 notesExercise #10 notes
Exercise #10 notes
 
X2 T07 02 transformations (2011)
X2 T07 02 transformations (2011)X2 T07 02 transformations (2011)
X2 T07 02 transformations (2011)
 
calculo vectorial
calculo vectorialcalculo vectorial
calculo vectorial
 
Mat 128 11 3
Mat 128 11 3Mat 128 11 3
Mat 128 11 3
 
X2 T04 03 cuve sketching - addition, subtraction, multiplication and division
X2 T04 03 cuve sketching - addition, subtraction,  multiplication and divisionX2 T04 03 cuve sketching - addition, subtraction,  multiplication and division
X2 T04 03 cuve sketching - addition, subtraction, multiplication and division
 
Figures
FiguresFigures
Figures
 
Cg
CgCg
Cg
 
Formula List Math 1230
Formula List Math 1230Formula List Math 1230
Formula List Math 1230
 
X2 T04 04 curve sketching - reciprocal functions
X2 T04 04 curve sketching - reciprocal functionsX2 T04 04 curve sketching - reciprocal functions
X2 T04 04 curve sketching - reciprocal functions
 

Similaire à 11X1 T14 04 areas

11 x1 t16 04 areas (2012)
11 x1 t16 04 areas (2012)11 x1 t16 04 areas (2012)
11 x1 t16 04 areas (2012)Nigel Simmons
 
Calculus cheat sheet_integrals
Calculus cheat sheet_integralsCalculus cheat sheet_integrals
Calculus cheat sheet_integralsUrbanX4
 
X2 t07 02 transformations (2012)
X2 t07 02 transformations (2012)X2 t07 02 transformations (2012)
X2 t07 02 transformations (2012)Nigel Simmons
 
Cea0001 ppt project
Cea0001 ppt projectCea0001 ppt project
Cea0001 ppt projectcea0001
 
Common derivatives integrals_reduced
Common derivatives integrals_reducedCommon derivatives integrals_reduced
Common derivatives integrals_reducedKyro Fitkry
 
The Definite Integral
The Definite IntegralThe Definite Integral
The Definite IntegralSilvius
 
584 fundamental theorem of calculus
584 fundamental theorem of calculus584 fundamental theorem of calculus
584 fundamental theorem of calculusgoldenratio618
 
Dsp U Lec07 Realization Of Discrete Time Systems
Dsp U   Lec07 Realization Of Discrete Time SystemsDsp U   Lec07 Realization Of Discrete Time Systems
Dsp U Lec07 Realization Of Discrete Time Systemstaha25
 
Emat 213 study guide
Emat 213 study guideEmat 213 study guide
Emat 213 study guideakabaka12
 
[4] num integration
[4] num integration[4] num integration
[4] num integrationikhulsys
 
Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...
Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...
Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...Mohd Paub
 
X2 T07 03 addition, subtraction, multiplication & division (2011)
X2 T07 03 addition, subtraction,  multiplication & division (2011)X2 T07 03 addition, subtraction,  multiplication & division (2011)
X2 T07 03 addition, subtraction, multiplication & division (2011)Nigel Simmons
 
X2 t07 03 addition, subtraction, multiplication & division (2012)
X2 t07 03 addition, subtraction,  multiplication & division (2012)X2 t07 03 addition, subtraction,  multiplication & division (2012)
X2 t07 03 addition, subtraction, multiplication & division (2012)Nigel Simmons
 
Areas of bounded regions
Areas of bounded regionsAreas of bounded regions
Areas of bounded regionsHimani Asija
 

Similaire à 11X1 T14 04 areas (20)

11 x1 t16 04 areas (2012)
11 x1 t16 04 areas (2012)11 x1 t16 04 areas (2012)
11 x1 t16 04 areas (2012)
 
Integration. area undera curve
Integration. area undera curveIntegration. area undera curve
Integration. area undera curve
 
Calculus cheat sheet_integrals
Calculus cheat sheet_integralsCalculus cheat sheet_integrals
Calculus cheat sheet_integrals
 
X2 t07 02 transformations (2012)
X2 t07 02 transformations (2012)X2 t07 02 transformations (2012)
X2 t07 02 transformations (2012)
 
Business math
Business mathBusiness math
Business math
 
Cea0001 ppt project
Cea0001 ppt projectCea0001 ppt project
Cea0001 ppt project
 
Common derivatives integrals_reduced
Common derivatives integrals_reducedCommon derivatives integrals_reduced
Common derivatives integrals_reduced
 
The Definite Integral
The Definite IntegralThe Definite Integral
The Definite Integral
 
584 fundamental theorem of calculus
584 fundamental theorem of calculus584 fundamental theorem of calculus
584 fundamental theorem of calculus
 
Dsp U Lec07 Realization Of Discrete Time Systems
Dsp U   Lec07 Realization Of Discrete Time SystemsDsp U   Lec07 Realization Of Discrete Time Systems
Dsp U Lec07 Realization Of Discrete Time Systems
 
Derivadas
DerivadasDerivadas
Derivadas
 
Exercise #8 notes
Exercise #8 notesExercise #8 notes
Exercise #8 notes
 
Emat 213 study guide
Emat 213 study guideEmat 213 study guide
Emat 213 study guide
 
[4] num integration
[4] num integration[4] num integration
[4] num integration
 
Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...
Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...
Comparison Of Dengue Cases Between Chosen District In Selangor By Using Fouri...
 
X2 T07 03 addition, subtraction, multiplication & division (2011)
X2 T07 03 addition, subtraction,  multiplication & division (2011)X2 T07 03 addition, subtraction,  multiplication & division (2011)
X2 T07 03 addition, subtraction, multiplication & division (2011)
 
Cs 601
Cs 601Cs 601
Cs 601
 
Exercise #11 notes
Exercise #11 notesExercise #11 notes
Exercise #11 notes
 
X2 t07 03 addition, subtraction, multiplication & division (2012)
X2 t07 03 addition, subtraction,  multiplication & division (2012)X2 t07 03 addition, subtraction,  multiplication & division (2012)
X2 t07 03 addition, subtraction, multiplication & division (2012)
 
Areas of bounded regions
Areas of bounded regionsAreas of bounded regions
Areas of bounded regions
 

Plus de Nigel Simmons

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATENigel 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
 
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
 
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
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)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
 
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 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)Nigel Simmons
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (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
 

Plus de 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)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (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 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (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)
 

Dernier

The basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxThe basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxheathfieldcps1
 
PSYCHIATRIC History collection FORMAT.pptx
PSYCHIATRIC   History collection FORMAT.pptxPSYCHIATRIC   History collection FORMAT.pptx
PSYCHIATRIC History collection FORMAT.pptxPoojaSen20
 
Presiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha electionsPresiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha electionsanshu789521
 
MENTAL STATUS EXAMINATION format.docx
MENTAL     STATUS EXAMINATION format.docxMENTAL     STATUS EXAMINATION format.docx
MENTAL STATUS EXAMINATION format.docxPoojaSen20
 
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...EduSkills OECD
 
Concept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.CompdfConcept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.CompdfUmakantAnnand
 
Hybridoma Technology ( Production , Purification , and Application )
Hybridoma Technology  ( Production , Purification , and Application  ) Hybridoma Technology  ( Production , Purification , and Application  )
Hybridoma Technology ( Production , Purification , and Application ) Sakshi Ghasle
 
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdfBASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdfSoniaTolstoy
 
“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...
“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...
“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...Marc Dusseiller Dusjagr
 
Alper Gobel In Media Res Media Component
Alper Gobel In Media Res Media ComponentAlper Gobel In Media Res Media Component
Alper Gobel In Media Res Media ComponentInMediaRes1
 
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxContemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxRoyAbrique
 
CARE OF CHILD IN INCUBATOR..........pptx
CARE OF CHILD IN INCUBATOR..........pptxCARE OF CHILD IN INCUBATOR..........pptx
CARE OF CHILD IN INCUBATOR..........pptxGaneshChakor2
 
URLs and Routing in the Odoo 17 Website App
URLs and Routing in the Odoo 17 Website AppURLs and Routing in the Odoo 17 Website App
URLs and Routing in the Odoo 17 Website AppCeline George
 
Accessible design: Minimum effort, maximum impact
Accessible design: Minimum effort, maximum impactAccessible design: Minimum effort, maximum impact
Accessible design: Minimum effort, maximum impactdawncurless
 
Sanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfSanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfsanyamsingh5019
 
Introduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxIntroduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxpboyjonauth
 
Measures of Central Tendency: Mean, Median and Mode
Measures of Central Tendency: Mean, Median and ModeMeasures of Central Tendency: Mean, Median and Mode
Measures of Central Tendency: Mean, Median and ModeThiyagu K
 
Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)eniolaolutunde
 
The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13Steve Thomason
 

Dernier (20)

Código Creativo y Arte de Software | Unidad 1
Código Creativo y Arte de Software | Unidad 1Código Creativo y Arte de Software | Unidad 1
Código Creativo y Arte de Software | Unidad 1
 
The basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxThe basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptx
 
PSYCHIATRIC History collection FORMAT.pptx
PSYCHIATRIC   History collection FORMAT.pptxPSYCHIATRIC   History collection FORMAT.pptx
PSYCHIATRIC History collection FORMAT.pptx
 
Presiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha electionsPresiding Officer Training module 2024 lok sabha elections
Presiding Officer Training module 2024 lok sabha elections
 
MENTAL STATUS EXAMINATION format.docx
MENTAL     STATUS EXAMINATION format.docxMENTAL     STATUS EXAMINATION format.docx
MENTAL STATUS EXAMINATION format.docx
 
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
 
Concept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.CompdfConcept of Vouching. B.Com(Hons) /B.Compdf
Concept of Vouching. B.Com(Hons) /B.Compdf
 
Hybridoma Technology ( Production , Purification , and Application )
Hybridoma Technology  ( Production , Purification , and Application  ) Hybridoma Technology  ( Production , Purification , and Application  )
Hybridoma Technology ( Production , Purification , and Application )
 
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdfBASLIQ CURRENT LOOKBOOK  LOOKBOOK(1) (1).pdf
BASLIQ CURRENT LOOKBOOK LOOKBOOK(1) (1).pdf
 
“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...
“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...
“Oh GOSH! Reflecting on Hackteria's Collaborative Practices in a Global Do-It...
 
Alper Gobel In Media Res Media Component
Alper Gobel In Media Res Media ComponentAlper Gobel In Media Res Media Component
Alper Gobel In Media Res Media Component
 
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptxContemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
Contemporary philippine arts from the regions_PPT_Module_12 [Autosaved] (1).pptx
 
CARE OF CHILD IN INCUBATOR..........pptx
CARE OF CHILD IN INCUBATOR..........pptxCARE OF CHILD IN INCUBATOR..........pptx
CARE OF CHILD IN INCUBATOR..........pptx
 
URLs and Routing in the Odoo 17 Website App
URLs and Routing in the Odoo 17 Website AppURLs and Routing in the Odoo 17 Website App
URLs and Routing in the Odoo 17 Website App
 
Accessible design: Minimum effort, maximum impact
Accessible design: Minimum effort, maximum impactAccessible design: Minimum effort, maximum impact
Accessible design: Minimum effort, maximum impact
 
Sanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdfSanyam Choudhary Chemistry practical.pdf
Sanyam Choudhary Chemistry practical.pdf
 
Introduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptxIntroduction to AI in Higher Education_draft.pptx
Introduction to AI in Higher Education_draft.pptx
 
Measures of Central Tendency: Mean, Median and Mode
Measures of Central Tendency: Mean, Median and ModeMeasures of Central Tendency: Mean, Median and Mode
Measures of Central Tendency: Mean, Median and Mode
 
Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)Software Engineering Methodologies (overview)
Software Engineering Methodologies (overview)
 
The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13The Most Excellent Way | 1 Corinthians 13
The Most Excellent Way | 1 Corinthians 13
 

11X1 T14 04 areas

  • 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
  • 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 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*