I need help with this and I need it fast!! (1 Viewer)

[JimMorrison]

Just a maths question....



Find the indefinite integral:

S (x^2 + x + 3) / 3x^-5 .dx


I'm just stumbling heaps.... any help would be greatly appreciated!

 

[jm1234567890]

http://integrals.wolfram.com/

 

[hiphophorray123]

woah just goes to show how quickly you can forget anything you learnt in 2 unit maths lol, if this was a few months ago i would of helped you no probs

 

[jm1234567890]

woah just goes to show how quickly you can forget anything you learnt in 2 unit maths lol, if this was a few months ago i would of helped you no probs

..... that's a pretty basic integral, there aren't even any tricks.

i think you should look at it again.

 

[katietheskatie]

that's what i was thinking .. i used to be able to do those no worries, but i just looked at that and was like 'wtf?'

 

[mr_brightside]

*whispers*

the answer is (a)

 

[JimMorrison]

holy crap that site is amazing!!!

the only problem is that it doesn't show working

 

[nick1048]

S (x^2 + x + 3)/(3x^-5)dx
= S 1/(3x^-5) . (x^2 + x + 3) dx
= S 3x^5(x^2 + x + 3)
= S 3x^7 + 3x^6 + 9x^5

= (3x^8/8) + (3x^7/7) + (9x^6/6)

 

[hiphophorray123]

..... that's a pretty basic integral, there aren't even any tricks.

i think you should look at it again.

yeh i should have looked at it again lol now i feel stupid

 

[nick1048]

yea that site is cool tho simplifies it nicely

 

[JimMorrison]

thanks heaps... the only thing is this is a major mindf"""" for me at the moment...

i copied the question incorrectly.... the same thing would apply with x^2 cancelling in this instance.....

S (x^2 + x + 3) / 3x^5 .dx (note the 5 is positive)

wouldn't it?

 

[JimMorrison]

wait, i just confused myself again........ the problem i had in the first place was i thought you couldn't bring the 3 up with the x^5??


arrrrrrrgh

 

[nick1048]

u get a totally different answer...

 

[war_ali]

= S 1/(3x^-5) . (x^2 + x + 3) dx
= S 3x^5(x^2 + x + 3)

I dont see how you can draw the 3 up to the nominator? thats like changing 1/3x to 3x^-1 instead of x^-1 / 3

 

[nick1048]

1/3x^-5 = 1/1/3x^5

= 1/1 divided by 1/3x^5

= 1/1 . 3x^5

= 3x^5

 

[JimMorrison]

= (3x^8/8) + (3x^7/7) + (9x^6/6)

isn't the answer i get from the integrals site??
(and the order of operations isn't wrong because i have tried multiple ways)

 

[nick1048]

physically do the division and clean up the algebra...

i.e. divide thru by 9...

= (x^6/6) + (x^7/21) + (x^8/24)

 

[war_ali]

can you show me that by doing 1 / (2x^-2)

 

[reincarnated]

im special

 

[nick1048]

can you show me that by doing 1 / (2x^-2)

1/(2x^-2)

You know that anything to the power of a negative means you can do this: a^-2 = 1/a^2

therefore 1 / [ 1 / (2x^2) ]

this is just a division of fractions:

1/1 divided by 1/2x^2

= 1/1 . 2x^2

= 2x^2