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Program Latih Tubi SPM Matematik Tambahan - Get as PowerPoint

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					TOPIC: INTEGRATIONS
                               dy                 √ K1
1. (a) y = k(x – 1)3      = 3k(x – 1)2(1)
                               dx
                                               dy
                      Given at point A, x = 3,    = 24
                                                         dx
        3k(x – 1)2 = 24;√ M1
         3k(3 – 1)2 = 24  k = 2 √ A1
   (b)(i) y = 2(x – 1) 3 ; On x-axis, y = 0
          0 = 2(x – 1) 3; x = 1
                                       3
          Area shaded region, P = 1     2( x 1)3 d x √ K1
                                3
             2 ( x- 1 )   4
                                             ( 3- 1 )4 
          =                       √   K1 =                √ M1
            
                 4            1
                                                2      

                                           = 8 √ A1
(b)(ii) Volume generated, R
            1
      = π 0[2( x- 1)3]2 d x
            1
          π 0 4( x - 1)6 d x
      =
                           1
             ( x - 1) 
                       7
          4π 
                  7
                              √ K1
      =               0

                 ( - 1 )7 
          4π 0               √ M1
                    7     
      =
          4π
      =           √ A1
           7
2.                                                         √ M1
     (a)   Area shaded region = Area under curve – Area Δ under straight line

             1                 1                  9
                                           
            0 ( 4y - y ) d y 2 ( k ) ( 4k k )  2 √ K1
                         2                   2




       4 y2   y3 k          1       9
√ K1 [         ]0 [ 2 k2  k 3]    √ K1
        2     3             2       2
       4 k2 k 3           1 3    9
√ M1 [         ] [ 2 k  k ] 
                        2
                                   √ M1
        2    3            2      2
            k3   9
               
            6    2

           k = 3 √ A1
                                               y
(b) V =  0 ( 4 y - y ) dy
              4          2   2


                                               4

           π 0 ( 16y  8y3  y 4 ) d y √ K1
              4      2
       =                                           x
                                               0
                  3      4       5
             16y 8y  y 4
       =   π[       ]0             √ M1
              3   4  5

         512π
       =              √ A1
          15

				
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