# Program Latih Tubi SPM Matematik Tambahan - Get as PowerPoint by nklye

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• pg 1
```									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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