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help on integration question- C3

the integration of (2x+a)^3 is equals to 90 with the limits a and 0. I need to find what a is.

I integrated the function and got (2x+a)^4/8 and the limits are a and 0. Substituted a and 0 into the integral giving ((3a)^4/8)-(a/8) = 90
then multiplied 3a by power of 4 giving :
(81a^4 / 8) - (a/8) = 90
brang the 8 to the other side giving:
81a^4-a=720
I then tried 2 solving for a numerous of ways and they all leaded me to the wrong answer so Im pretty sure something went wrong already...
so how would i find a?
Reply 1
Original post by TwlilightLoz
the integration of (2x+a)^3 is equals to 90 with the limits a and 0. I need to find what a is.

I integrated the function and got (2x+a)^4/8 and the limits are a and 0. Substituted a and 0 into the integral giving ((3a)^4/8)-(a/8) = 90
then multiplied 3a by power of 4 giving :
(81a^4 / 8) - (a/8) = 90
brang the 8 to the other side giving:
81a^4-a=720
I then tried 2 solving for a numerous of ways and they all leaded me to the wrong answer so Im pretty sure something went wrong already...
so how would i find a?


Where does your a8\dfrac{a}{8} come from

Why is it not a48\dfrac{a^4}{8}
Reply 2
Ohh yeah, i forgot to add that.

EDIT = tried solving it but couldn't. I have 81a^4 - a^4 = 720 --> 80a^4=720 --> a^4=9 --> a=4sqrt9 --> answer is wrong, it should be sqrt3
Reply 3
Original post by TwlilightLoz
Ohh yeah, i forgot to add that.

EDIT = tried solving it but couldn't. I have 81a^4 - a^4 = 720 --> 80a^4=720 --> a^4=9 --> a=4sqrt9 --> answer is wrong, it should be sqrt3


a4=9a^4=9 is correct so a=91/4 a=9^{1/4}.
Reply 4
Original post by BabyMaths
a4=9a^4=9 is correct so a=91/4 a=9^{1/4}.


ohhh, i didnt even notice that, thanks for the help
Reply 5
Original post by TwlilightLoz
Ohh yeah, i forgot to add that.

EDIT = tried solving it but couldn't. I have 81a^4 - a^4 = 720 --> 80a^4=720 --> a^4=9 --> a=4sqrt9 --> answer is wrong, it should be sqrt3


49=3^4\sqrt9 = \sqrt3
Reply 6
Original post by TenOfThem
49=3^4\sqrt9 = \sqrt3


Well in the book it does say sqrt3 but the question didnt specifiy on how they want the answer to be written. If this was an exam would I loose a mark for writing 9^1/4 rather than sqrt3?
Reply 7
Original post by TwlilightLoz
If this was an exam would I loose a mark for writing 9^1/4 rather than sqrt3?


In general you should try to give the answer in its simplest form.

I wasn't offering 91/49^{1/4} as the answer, just as a step in the right direction.

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