integration Q (1 Viewer)

[pLuvia]

Evaluate the integral

∫(2x^3/2 - 1)^1/3 √x dx

can someone help me on this please?

 

[Dreamerish*~]

Evaluate the integral

∫(2x^3/2 - 1)^1/3 √x dx

can someone help me on this please?

Ok... um, does the question look like this?

∫(2x^3/₂ - 1)^1/₃ √x dx

OR

∫(2x^3/₂ - 1)^√x/₃?

 

[haboozin]

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]

Ok... um, does the question look like this?

∫(2x^3/₂ - 1)^1/₃ √x dx

OR

∫(2x^3/₂ - 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)

 

[KFunk]

Evaluate the integral

∫(2x^3/2 - 1)^1/3 √x dx

can someone help me on this please?

if f(x) = 2x^3/2 -1

f '(x) = 3x^1/2

int = 1/3.∫[f(x)]^1/3[f'(x)] dx so let u = f(x)

int = 1/3.∫ u^1/3 du = 1/4.u^4/3 +C

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

^ As above, .

 

[pLuvia]

Ok... um, does the question look like this?

∫(2x^3/₂ - 1)^1/₃ √x dx

OR

∫(2x^3/₂ - 1)^√x/₃?

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

 

[Antwan23q]

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

 

[pLuvia]

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..

 

[pLuvia]

this is what ive got so far


∫(2x^3/₂ - 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]

∫(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]

∫(2x^3/₂ - 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.

 

[haboozin]

∫(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]

this is what ive got so far


∫(2x^3/₂ - 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

 

[pLuvia]

omg.. far out i wrote out the wrong question.... my bad

heres the real question sorry

∫x(3x-2)^3 dx

heres the one that i was trying to ask sorry

 

[rama_v]

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]

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.

 

[pLuvia]

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??

 

[richz]

huh?? didnt they cancel??

 

[pLuvia]

huh?? didnt they cancel??

how did it cancel??.. sorry i only started learning integration last saturday... it shouldnt cancel should it?

 

[Trev]

how did it cancel??.. sorry i only started learning integration last saturday... it shouldnt cancel should it?

∫(2x^3/2 - 1)^1/3 √x dx

Let u = 2x^3/2 - 1
du/dx = 3√x
du/3 = √x.dx
∴ ∫(2x^3/2 - 1)^1/3 √x dx = ∫(u)^1/3(du/3) etc etc.
Do you understand it now?