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Integration question ...please help!
Started by Azie, May 12 2007 12:20 PM
5 replies to this topic
#1
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Posted 12 May 2007 - 12:20 PM
its question 7 in paper 1 of the 2004 past papers ...i've tried but i cant seem 2 get the rite answer ...can any1 please show me how 2 work it out as simply as possible with as much workin ...thanks a lot!
#2
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Posted 12 May 2007 - 03:36 PM
Sorry I tried to use LaTex but I just cant understand it!
(4x+1)^1/2
(4x+1)^3/2 /6
(9^3/2 /6 ) - ( 1^3/2 /6)
=27/6 - 1/6 = 26/6
(4x+1)^1/2
(4x+1)^3/2 /6
(9^3/2 /6 ) - ( 1^3/2 /6)
=27/6 - 1/6 = 26/6
#3
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Posted 12 May 2007 - 08:01 PM
how did u get the "/6" part in the 2nd line of working?
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Posted 13 May 2007 - 04:53 PM
QUOTE(celticfcz @ May 12 2007, 04:36 PM) 
Sorry I tried to use LaTex but I just cant understand it!
(4x+1)^1/2
(4x+1)^3/2 /6
(9^3/2 /6 ) - ( 1^3/2 /6)
=27/6 - 1/6 = 26/6
(4x+1)^1/2
(4x+1)^3/2 /6
(9^3/2 /6 ) - ( 1^3/2 /6)
=27/6 - 1/6 = 26/6
Sorry - missed a step.
When you integrate you add one on to the power then take it to the bottom and divide it by that power - in further integration you multiply that by the middle bit differentiated (whats in the brackets) so that =4.
so 3/2 x 4 is at the bottom - you flip it ie take the 2 up whaich gives : 2(4x+1)^3/2 all divided by 3x4 which is 12.
then simply cancel the 2 and the 12 leaving you with 6.
Hope that helps.
#5
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Posted 13 May 2007 - 05:09 PM
yay! i finally got it, thanks a lot
#6
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Posted 13 May 2007 - 05:48 PM
no problem!
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