Maths Question
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elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
I have a question that I can't seem to do.

Can someone explain how to Integrate :



For the first one, it is just the root x + 2x over the x squared. I know with 1/x^2 it would be Ln, but what about with the root x + 2x.

As for the second one, the section with the v I don't know what it means.



Edited by elster on Monday 17th May 18:55

toasty

8,414 posts

250 months

Monday 17th May 2010
quotequote all
potato

elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
toasty said:
potato
I tried the magic 8 ball, I never thought about a potato. Good idea batman.

43034

2,971 posts

198 months

Monday 17th May 2010
quotequote all
Sorry I can't be of any help, but wondering is this A-level work or higher?

Ta.

elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
43034 said:
Sorry I can't be of any help, but wondering is this A-level work or higher?

Ta.
A-Level, as I'm doing foundation year at University.

MoleVision

999 posts

241 months

Monday 17th May 2010
quotequote all
elster said:
I have a question that I can't seem to do.

Can someone explain how to Integrate :



For the first one, it is just the root x + 2x over the x squared. I know with 1/x^2 it would be Ln, but what about with the root x + 2x.

As for the second one, the section with the v I don't know what it means.



Edited by elster on Monday 17th May 18:55
for the 1st one.. err no.6. you need to use proportional integration (i think - its called something like that) where you split it into 2 halves.. one half being 1/X^2.

The 2nd one , no.7 I would assume that you can treat anything that isnt 'squiggle' as a constant which makes it a lot easier.

someone will proabbly be along in a sec to tell me I've forgotten everything I learnt at school and Uni (which are both a few years ago now!)


james_gt3rs

4,816 posts

221 months

Monday 17th May 2010
quotequote all
You can re-write the equation as:

(x^-3/2) + (2x^-1).

Then it is easy to integrate from there.

I don't recognise the symbols on the second question though.

elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
MoleVision said:
elster said:
I have a question that I can't seem to do.

Can someone explain how to Integrate :



For the first one, it is just the root x + 2x over the x squared. I know with 1/x^2 it would be Ln, but what about with the root x + 2x.

As for the second one, the section with the v I don't know what it means.



Edited by elster on Monday 17th May 18:55
for the 1st one.. err no.6. you need to use proportional integration (i think - its called something like that) where you split it into 2 halves.. one half being 1/X^2.

The 2nd one , no.7 I would assume that you can treat anything that isnt 'squiggle' as a constant which makes it a lot easier.

someone will proabbly be along in a sec to tell me I've forgotten everything I learnt at school and Uni (which are both a few years ago now!)
Ah so the first one replace the top line with u, integrate in terms of u.

As for the second one, how would I do that?

james_gt3rs

4,816 posts

221 months

Monday 17th May 2010
quotequote all
Hang on, just checked Wolfram alpha, comes up with something different, so I may be wrong.

http://www.wolframalpha.com/input/?i=integrate+%28...

CobolMan

1,432 posts

237 months

Monday 17th May 2010
quotequote all
elster, for the first one split it up into 2 expressions, each of which is divided by x-squared. If I'm right, you should end up with
2 - √2 + 2ln2

Not sure about the second one, I'll see if I can figure out some pointers for you.

Edited to correct the √2 bit

Edited by CobolMan on Monday 17th May 20:37

navier_stokes

948 posts

229 months

Monday 17th May 2010
quotequote all
The second one you only have to integrate with respect to Zeta.... treat everything else as constants. It's easier than the first smile

Dixie68

3,091 posts

217 months

Monday 17th May 2010
quotequote all
43034 said:
Sorry I can't be of any help, but wondering is this A-level work or higher?

Ta.
Judging by the exam papers posted on PH the other day the GCSE paper would've said, "1+1=2, what does 1+1 equal"?

pugwash4x4

7,682 posts

251 months

Monday 17th May 2010
quotequote all
It was in an A-level module SWMBO took about 12 years ago- would be interested to know if it still was!

MoleVision

999 posts

241 months

Monday 17th May 2010
quotequote all
elster said:
Ah so the first one replace the top line with u, integrate in terms of u.
Yeh I think so, cant remember the procedure exactly... dont think expanding it really helps cos you still end up with having to integrate rootx/(x^2).


elster said:
As for the second one, how would I do that?
well you are integrating with respect to squiggle or "Zeta" as some call it..(Z when written below)... so treat the x^2 and the av as simple constants then its basic integration.

the x^2 then integrates to Z(x^2)
the 2/Z integrates to 2lnZ or something (thats a guess - you will need to check)
the avZ integrates to (av(Z^2))/2

I think.
Again, someone may want to tell me I'm wrong (or hopefully right!)

CobolMan

1,432 posts

237 months

Monday 17th May 2010
quotequote all
I hope this comes out OK:


elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
CobolMan said:
I hope this comes out OK:

Ah so it isn't done by substitution.

It looks so easy now you have written it down like that. Between stage 2 and 3 is the bit I was unsure of.

Thanks.

elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
MoleVision said:
well you are integrating with respect to squiggle or "Zeta" as some call it..(Z when written below)... so treat the x^2 and the av as simple constants then its basic integration.

the x^2 then integrates to Z(x^2)
the 2/Z integrates to 2lnZ or something (thats a guess - you will need to check)
the avZ integrates to (av(Z^2))/2

I think.
Again, someone may want to tell me I'm wrong (or hopefully right!)
That is all fine, I was just unsure what the alpha v meant. If I just class it as alpha-v times zeta.

So when you are integrating in terms of zeta, you treat and other letters as zeta^0.

CobolMan

1,432 posts

237 months

Monday 17th May 2010
quotequote all
HTH elster - you wouldn't believe the 'fun' I had getting the formatting correct though wink

elster

Original Poster:

17,517 posts

240 months

Monday 17th May 2010
quotequote all
CobolMan said:
HTH elster - you wouldn't believe the 'fun' I had getting the formatting correct though wink
What do you use to write the formula?


Thanks all for the help as well guys.

CobolMan

1,432 posts

237 months

Monday 17th May 2010
quotequote all
elster said:
CobolMan said:
HTH elster - you wouldn't believe the 'fun' I had getting the formatting correct though wink
What do you use to write the formula?


Thanks all for the help as well guys.
I used the formula editor in a NeoOffice word processing document then took a screen grab of the file onto my desktop, saved it as a .png in Preview and then uploaded it. Good luck with your course.