Basic intergration

  • Thread starter jamesd2008
  • Start date
64
0

Main Question or Discussion Point

Hi

could someone explain to me why, if I integrate dv/v between V2 and V1 the result is nastural log of V2/V1?

Thanks
 

Answers and Replies

Cyosis
Homework Helper
1,495
0
Is your question why the primitive of 1/v is log v or are you wondering why you get log v2/v1 instead of log v2 - log v1? Anyhow [itex]\log v_2-\log v_1=\log v_2/v_1[/itex].
 
64
0
Yes thanks Cyosis!

Missed the fact that logV2-logV1=LogV2/V1

Thank you very much for your insight!
james
 
64
0
Also is not that 1/v.dv between v2 and v1 = [log|v2|-log|v1|] between v2 and v1? and that the dv is ignored?
 
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
12,038
128
The dv is not ignored, it is necessary in an integral to specify the variable of integration.
 
lavinia
Science Advisor
Gold Member
3,079
538
Also is not that 1/v.dv between v2 and v1 = [log|v2|-log|v1|] between v2 and v1? and that the dv is ignored?
can you restate this more clearly? I don't get what you are asking.
 
could someone explain to me why, if I integrate dv/v between V2 and V1 the result is nastural log of V2/V1?
[tex]\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{.}[/tex]
Also is not that 1/v.dv between v2 and v1 = [log|v2|-log|v1|] between v2 and v1? and that the dv is ignored?
[tex]\text{Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}[/tex]
 
Last edited:
64
0
Hi, I think what I'm getting confused about, is that for entropy the change in entropy is ds=dQ/T. Integrating this gives s2-s1=the integral of dQ/T between 2 and 1. So are you saying that the change in Q is now just there to specify the variable of integration? Sorry id this all sounds confusing.
 
48
1
It does, apology accepted :)
 
24
0
integral (dv / v) = integral (1/v)*dv = integral (1/v) dv = ln|v|

since our integral is between V2 and V1, we do ln|V2| - ln|V1| = ln|V2/V1|
 
...
since our integral is between V2 and V1, we do ln|V2| - ln|V1| = ln|V2/V1|
[tex]\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{. ...}[/tex]
[tex]\text{ ... Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}[/tex]
(As before.)
 
64
0
Thanks everyone for there help.
 
Char. Limit
Gold Member
1,198
12
[tex]\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{.}[/tex]

[tex]\text{Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}[/tex]
If you integrate from v2 to v1, yes, that is correct. I was under the impression that e was integrating from v1 to v2. The word "between" creates an ambiguity there, I suppose.
 

Related Threads for: Basic intergration

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
6
Views
2K
Top