Integration by substitution q (1 Viewer)

[x.Exhaust.x]

I'm stuck . Could anyone possibly point out the flaws in my working out if there is any? If not, how would I continue?

Thanks in advance.

 

[Trebla]

You sure the question is correct? I think it should say 12x²√(3 - x³)...for a substitution question

 

[Timothy.Siu]

even if u did it correct, u still have and x in the question so u cannot integrate it with what u've done
and i agree with trebla, the question is a bit dodgey, i subbed in x=(3cos²a)^1/3
but it got a bit messy, so i think there is some error in the question

 

[x.Exhaust.x]

[FONT='Calibri','sans-serif']So far I've got (see attachment).[/FONT]
[FONT='Calibri','sans-serif'][/FONT]
[FONT='Calibri','sans-serif']Is there any possible way to integrate that further? This exact question was in my 3U exam, worth 4 or 5 marks (either two..).[/FONT][FONT='Calibri','sans-serif']
[/FONT]

 

[lyounamu]

[FONT='Calibri','sans-serif']So far I've got (see attachment).[/font]
[FONT='Calibri','sans-serif'][/font]
[FONT='Calibri','sans-serif']Is there any possible way to integrate that further? This exact question was in my 3U exam, worth 4 or 5 marks (either two..).[/font][FONT='Calibri','sans-serif']
[/font]

can you give us the original question?

Or is the original attachment (in the OP) one the original question?

some people have already noted that the question may have been flawed.

 

[x.Exhaust.x]

I'm pretty sure that is the original question Namu...My friend got 2root2 as an answer, (root 8), but I'm not sure if that's correct...

So with the current attachment, is there any possible way to integrate

 

[lyounamu]

Yeah, I think you can do it. It will get extremely messy.

The purpose of using a substitution method is to get the question simplified, not overally expanded. I really doubt that working that part out will get you a right answer.

 

[lolokay]

http://integrals.wolfram.com/index.jsp?expr=12x*sqrt(3-x^3)&random=false

 

[lyounamu]

I got you the integral for you:

looking over this and lolokay's (which is exactly same as mine), i doubt that was your question. I mean, they wouldn't have asked you that.

 

[lyounamu]

I did the question according to trebla and mine looks like this:

I (12x^2 . sqrt(3-x^3) dx terminals are x=1 and x=0
let u = 3-x^3
du/dx = -3x^2

now

-4 I (-3x^2 . sqrt (3-x^3) dx
then
-4 I (sqrt(u) du terminals are u=2 and u=3
then

-4 [2/3 . u^(3/2)] and terminals are u=2 and u=3

then it should get you:

-8/3 (2^3/2 - 3^3/2)
= 6.31393412... = 6.31 to 2 d.p.

 

[x.Exhaust.x]

Ah thanks anyway. I'll talk to my teacher. She may have wrote it wrong...