Last Maths qns
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how u integrate [ x^3 - 10/x^2] dx
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(1/4) x^4 + (10/x) + c where c is an arbitrary constant.
Don't think hard enough for me to show working....
EDIT: Paiseh, corrected answer....
EDIT2: too tired already.... -
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you are uniqueOriginally posted by starblue: -
x^4/4 + 10/xOriginally posted by FadeToBlack:how u integrate [ x^3 - 10/x^2] dx -
i thought is (10/-1) x^-1? how u get 3?Originally posted by CoolMyth:1/4 x^4 - (10/3)x^3 + c where c is an arbitrary constant. -
i got your ans oso, but if i give u the bounderies 4 and 2, can u work out the ans? coz i got a diff ans from the TYS AGAIN!Originally posted by ditzy:x^4/4 + 10/x
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correct answer, I repeat is:Originally posted by FadeToBlack:i thought is (10/-1) x^-1? how u get 3?
(1/4)x^4 + (10/x) + c where c is an arbitrary constant.
Just now didn't see the / clearly.....
EDIT: fatal mistake -
fadetoback is also maths major??!!
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i got this x^4/4 + 10/x + c same as ditzy, how u get 5/x?Originally posted by CoolMyth:correct answer, I repeat is:
(1/4)x^4 + (5/x) + c where c is an arbitrary constant.
Just now didn't see the / clearly..... -
paiseh, paiseh....made error....Originally posted by FadeToBlack:i got this x^4/4 + 10/x + c same as ditzy, how u get 5/x?
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[4^4/4 + 10/4] - [2^4/4 + 10/2]Originally posted by FadeToBlack:i got your ans oso, but if i give u the bounderies 4 and 2, can u work out the ans? coz i got a diff ans from the TYS AGAIN!
work out yourself lah. -
more of a maths failure...Originally posted by starblue:fadetoback is also maths major??!! -
its not 5/xOriginally posted by CoolMyth:correct answer, I repeat is:
(1/4)x^4 + (5/x) + c where c is an arbitrary constant.
Just now didn't see the / clearly.....
its 10/x -
"Originally posted by ditzy:[4^4/4 + 10/4] - [2^4/4 + 10/2]
work out yourself lah.
no u gimme ur ans, i wanna check if im wrong or not.. and oso coolmyth got a diff ans.. -
Full working:
|(2,4) x^3 - 10/x^2 dx
= [(1/4)x^4 + 10/x](2,4)
= (64 + 2.5) - (4 + 5)
= 57.5 -
ok nvm, stayOriginally posted by CoolMyth:paiseh, paiseh....made error.... 
.... given bounderies 4 and 2 u can help me work ans out? -
Have abit of faith in yourself can boh? You correct lah, the book wrong lah. I lazy to take out calculator now k?
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oooooh i noe where i wrong liao, i sub in 4 first.... but can i asked a qns?Originally posted by CoolMyth:Full working:
|(2,4) x^3 - 10/x^2 dx
= [(1/4)x^4 + 10/x](2,4)
= (64 + 2.5) - (4 + 5)
= 57.5
why u sub 2 first ar? isn't 4 a bigger number? -
57.5Originally posted by FadeToBlack:i got your ans oso, but if i give u the bounderies 4 and 2, can u work out the ans? coz i got a diff ans from the TYS AGAIN! -
you sub in the upper limit of the boundary.Originally posted by FadeToBlack:oooooh i noe where i wrong liao, i sub in 4 first.... but can i asked a qns?
why u sub 2 first ar? isn't 4 a bigger number?
then minus the lower limit of the boundary
in this case...
4 is the upper limit...
2 is the lower limit... -
[64 + 5/2] - [4 + 5]
Simple? Satisfied? -
when I say [f(x)](2,4), it means f(4) - f(2)....Originally posted by FadeToBlack:oooooh i noe where i wrong liao, i sub in 4 first.... but can i asked a qns?
why u sub 2 first ar? isn't 4 a bigger number?
you don't expect me to write and paste a pic here do you....? -
shit la im correct, pai seh pai seh, dunno wtf am i doing.... already got the steps all correct liao then last step calculate wrong... haix, tmr test die liao..Originally posted by CoolMyth:when I say [f(x)](2,4), it means f(4) - f(2)....
you don't expect me to write and paste a pic here do you....?