Basic intergration

1. May 17, 2010

jamesd2008

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

2. May 17, 2010

Cyosis

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 $\log v_2-\log v_1=\log v_2/v_1$.

3. May 17, 2010

jamesd2008

Yes thanks Cyosis!

Missed the fact that logV2-logV1=LogV2/V1

Thank you very much for your insight!
james

4. May 17, 2010

jamesd2008

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?

5. May 18, 2010

Redbelly98

Staff Emeritus
The dv is not ignored, it is necessary in an integral to specify the variable of integration.

6. May 18, 2010

lavinia

can you restate this more clearly? I don't get what you are asking.

7. May 18, 2010

Martin Rattigan

$$\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{.}$$
$$\text{Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}$$

Last edited: May 18, 2010
8. May 18, 2010

jamesd2008

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.

9. May 18, 2010

phizo

It does, apology accepted :)

10. May 24, 2010

luma

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|

11. May 25, 2010

Martin Rattigan

(As before.)

12. May 25, 2010

jamesd2008

Thanks everyone for there help.

13. May 25, 2010

Char. Limit

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.