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### #1Azie

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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!

### #2celticfcz

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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

### #3Azie

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Posted 12 May 2007 - 08:01 PM

how did u get the "/6" part in the 2nd line of working?

### #4celticfcz

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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

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.

### #5Azie

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Posted 13 May 2007 - 05:09 PM

yay! i finally got it, thanks a lot

### #6celticfcz

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Posted 13 May 2007 - 05:48 PM

no problem!

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