Integration (1 Viewer)

[Grey Council]

can't do this one:

Integrate:
x ( sqrt (3x - 1))

 

[nike33]

answers or hints?, anyway an easy way (not the best) is let u = 3x-1

 

[Grey Council]

I tried!

damn it, the answer at the back is weird. :-S

just tell me answer, and if its different from mine, i'll hafta ask/try again.

 

[CM_Tutor]

I get u = 3x - 1 giving an answer of 2(3x - 1)(9x + 2) * sqrt(3x - 1) / 135 + C, for some constant C.

 

[nike33]

yes i got that as well

 

[Giant Lobster]

answer at the back of ur book and urs prolly differ by a constant

just dy/dx ur answer to check

 

[George W. Bush]

While this question is on the table, a quick question for CM_Tutor if I may:

For a simple substitution question (assuming I don't want to do the above another way), must I write out all the steps e.g.
Let u = 3x-1, du = 3dx etc., or can I simply imply the use of a substitution?

 

[Affinity]

write the substitution (the u = 3x -1 part), the du = 3dx part is optional unless it's not obvious at all (ie you have to work it out on paper)

 

[George W. Bush]

but i mean, say the question was integrate sinx cos^4xdx. Could I simply write the answer or would it be necessary to write Let u = cosx, -u^4 etc.

 

[Affinity]

if it is find the integral, just write the answer down

if they specified a substitution then show some steps

 

[DcM]

arghh y didnt i get that..

did pplz get:

[ 2(3x-1)^5/2 + 2(3x-1)^3/2 ] / 135 + C

b4 simplifying ie..

 

[CM_Tutor]

Originally posted by DcM
arghh y didnt i get that..

did pplz get:

[ 2(3x-1)^5/2 + 2(3x-1)^3/2 ] / 135 + C

b4 simplifying ie..

No, because you have made an error in the step to get the 135. It should be:

[ 2 * 3 * (3x-1)^5/2 + 2 * 5 * (3x-1)^3/2 ] / 135 + C

 

[Teoh]

Whoa
I just realised, I have no idea how to do this???

}x sqrt (3x - 1) dx

let u = 3x -1

du/dx = 3 > dx = du/3

} x (sqrt u) du/3


Is this anywhere near right...? Then what to you do?

 

[CM_Tutor]

Originally posted by Teoh
Is this anywhere near right...? Then what to you do?

You are on the right track. You need to replace the x in the integral with (u + 1) / 3, so that you have an expression just in u, integrate it with respect to u, and then change the result back to using the variable x.

 

[Teoh]

the x in the integral with (u + 1) / 3

don't get it

 

[Xayma]

u=3x-1

3x=u+1
x=(u+1)/3

So your intergral will look like

/
| (u+3u^1/2+1)</sup>/3
/

 

[CM_Tutor]

Xayma, I think you have made a mistake in your expansion / substituiton.

In any case, here is the full answer. Let I = int x * sqrt(3x - 1) dx

Now, let u = 3x - 1, and use the Reverse Chain Rule.
So, du = 3dx, and dx = du / 3
Also, if u = 3x - 1, then x = (u + 1) / 3

So, I = int [(u + 1) / 3] * sqrt(u) du / 3
= (1 / 9) * int (u + 1)sqrt(u) du
= (1 / 9) * int u * sqrt(u) + sqrt(u) du
= (1 / 9) * int u¹u^1/2 + u^1/2 du
= (1 / 9) * int u^3/2 + u^1/2 du
= (1 / 9) * [u^5/2 / (5/2) + u^3/2 / (3/2)] + C, for some constant C
= (1 / 9) * [2u^2+1/2 / 5 + 2u^1+1/2 / 3] + C
= (1 / 9) * [2u²sqrt(u) / 5 + 2u * sqrt(u) / 3] + C
= (2u * sqrt(u) / 9) * [(u / 5) + (1 / 3)] + C
= [2u * sqrt(u) / (9 * 3 * 5)](3u + 5) + C
= 2u(3u + 5) * sqrt(u) / 135 + C

But u = 3x - 1

So, I = 2(3x - 1)[3(3x - 1) + 5] * sqrt(3x - 1) / 135 + C
= 2(3x - 1)(9x + 2) * sqrt(3x - 1) / 135 + C, as stated above.

 

[Teoh]

Damn I feel guilty, all that type, and for such a relatively easy question...tsk tsk

u = 3x - 1
x = (1 + u)/3

How easy is that? Gawd its ridiculous the things that u can't see sometimes...

 

[Xayma]

Originally posted by CM_Tutor
Xayma, I think you have made a mistake in your expansion / substituiton.

Oh yeah, it was a times wasnt it, opps, but I was only doing it from memory and trying to explain the x=(1+u)/3

 

[Grey Council]

umm, did someone give any thought to the question i posted up earlier?

when you hafta prove that one curve is higher than another curve, and then the question asks you to prove that the integral of one of the curves is more than a certain number but less than another.
Inequations and integration, i think the topic is.

where abouts in a paper would that appear?