integration Q (1 Viewer)

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pLuvia

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Evaluate the integral

∫(2x^3/2 - 1)^1/3 √x dx

can someone help me on this please? :p:)
 

Dreamerish*~

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kadlil said:
Evaluate the integral

∫(2x^3/2 - 1)^1/3 √x dx

can someone help me on this please? :p:)
Ok... um, does the question look like this?

∫(2x3/2 - 1)1/3 √x dx

OR

∫(2x3/2 - 1)√x/3?
 

haboozin

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u = 2x^3/2 -1
u' = 3x^1/2

1/3 I u^1/3 du

= 1/4 u^4/3 + c

= 1/4 (2x^3/2 - 1)^4/3) + c
 

haboozin

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Dreamerish*~ said:
Ok... um, does the question look like this?

∫(2x3/2 - 1)1/3 √x dx

OR

∫(2x3/2 - 1)√x/3?

haha obviously the first one...
no one here coudl do the second one.. without having other knowledge outside the syllabous.


ps u will never get this question posted in the hsc anyways unless they give u the substitution (in 3unit that is)
 

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kadlil said:
Evaluate the integral

∫(2x^3/2 - 1)^1/3 √x dx

can someone help me on this please? :p:)
if f(x) = 2x<sup>3/2</sup> -1

f '(x) = 3x<sup>1/2</sup>

int = 1/3.&int;[f(x)]<sup>1/3</sup>[f'(x)] dx so let u = f(x)

int = 1/3.&int; u<sup>1/3</sup> du = 1/4.u<sup>4/3</sup> +C

= 1/4.(2x<sup>3/2</sup> -1)<sup>4/3</sup> +C

^ As above, :p.
 
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pLuvia

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Dreamerish*~ said:
Ok... um, does the question look like this?

∫(2x3/2 - 1)1/3 √x dx

OR

∫(2x3/2 - 1)√x/3?


∫(2x3/2 - 1)1/3 √x dx
this one sorry bout the writing lol.. i couldnt get the power fractions
 

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look at ya. someone posts an integration and u all flock like seagalls to do it. Yes k-funk im lookin at u. X (
 
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pLuvia

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haboozin said:
haha obviously the first one...
no one here coudl do the second one.. without having other knowledge outside the syllabous.


ps u will never get this question posted in the hsc anyways unless they give u the substitution (in 3unit that is)
oh umm.. this Q is a substitution Q.. sorry but your answers are wrong... i have the answers but i can't do the Question..
 
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pLuvia

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this is what ive got so far


∫(2x3/2 - 1)1/3 √x dx

Let u=3x-2
x=u+2/3
dx \ du/3

1/3 ∫(u+2)u^3 du/3

= 1/9 ∫u^4 + 2u^3 du

= 1/9(5u^5 + 8u^4) +C

but i think this is wrong :(
 

mynameisgone

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∫(2x3/2 - 1)^1/3 √x dx

let u=x^3/2
du = 3/2.x^1/2.dx

I = 2/3.∫(2u-1)^1/3.du

I = 2/3.[3/8.(2u-1)^4/3] + C

I = 6/24.(2u-1)^4/3 + C

I = 6/24.(2x^3/2-1)^4/3 + C
 

haboozin

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kadlil said:
∫(2x3/2 - 1)1/3 √x dx

hey, if your question was infact bracket times the sqaureroot of x then that is the answer.

it is possible that your formatting or the answer you have has formatting which is different to mine and kfunk's.

but i assure you if this is the integration you gave us then it is correct.

another reason why the question seams right (well what i think the question is anyways) is because it actually integrates easily.
 
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haboozin

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mynameisgone said:
∫(2x3/2 - 1)^1/3 √x dx

let u=x^3/2
du = 3/2.x^1/2.dx

I = 2/3.∫(2u-1)^1/3.du

I = 2/3.[3/8.(2u-1)^4/3] + C

I = 6/24.(2u-1)^4/3 + C

I = 6/24.(2x^3/2-1)^4/3 + C

6 is divisible by 24

1/4() + c

exact same as what i got..
 

haboozin

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kadlil said:
this is what ive got so far


∫(2x3/2 - 1)1/3 √x dx

Let u=3x-2
x=u+2/3
dx \ du/3

1/3 ∫(u+2)u^3 du/3

= 1/9 ∫u^4 + 2u^3 du

= 1/9(5u^5 + 8u^4) +C

but i think this is wrong :(


why did u think of 3x - 2 ??
what does that have to do with anything to do with this question..

the use of substitution is really just to make u see the steps since you are reversing your differentiation and some of the steps can get difficult.

eg if you are integrating x using the substitution fo u = x + 1 will just make your steps longer
 
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pLuvia

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omg.. far out i wrote out the wrong question.... :(:p my bad

heres the real question sorry

∫x(3x-2)^3 dx

heres the one that i was trying to ask sorry
 

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let u = 3x-2
du=3dx
(1/3) du = x dx
Int = (1/3) u^3 du
= (1/3) (u^4)/4 +C
= (1/3)((3x-2)^4)/4 + C
Im not sure if thats right im pretty tired right now...
 

haboozin

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rama_v said:
let u = 3x-2
du=3dx
(1/3) du = x dx
Int = (1/3) u^3 du
= (1/3) (u^4)/4 +C
= (1/3)((3x-2)^4)/4 + C
Im not sure if thats right im pretty tired right now...

looks plenty right to me.
 
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pLuvia

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rama_v said:
let u = 3x-2
du=3dx
(1/3) du = x dx
Int = (1/3) u^3 du
= (1/3) (u^4)/4 +C
= (1/3)((3x-2)^4)/4 + C
Im not sure if thats right im pretty tired right now...
but wat happened to the "x"

∫x(3x-2)^3 dx??
 
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pLuvia

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xrtzx said:
huh?? didnt they cancel??
how did it cancel??.. sorry i only started learning integration last saturday... it shouldnt cancel should it?
 

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kadlil said:
how did it cancel??.. sorry i only started learning integration last saturday... it shouldnt cancel should it?
∫(2x<sup>3/2</sup> - 1)<sup>1/3</sup> √x dx

Let u = 2x<sup>3/2</sup> - 1
du/dx = 3√x
du/3 = √x.dx
&there4; ∫(2x<sup>3/2</sup> - 1)<sup>1/3</sup> √x dx = ∫(u)<sup>1/3</sup>(du/3) etc etc.
Do you understand it now?
 

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