Explaining Basic Integration: dv/v, Natural Log of V2/V1

[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

 

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

 

[jamesd2008]

Yes thanks Cyosis!

Missed the fact that logV2-logV1=LogV2/V1

Thank you very much for your insight!
james

 

[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?

 

[Redbelly98]

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

 

[lavinia]

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.

 

[Martin Rattigan]

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

$$\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{.}$$

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?

$$\text{Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}$$

 

[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.

 

[phizo]

It does, apology accepted :)

 

[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|

 

[Martin Rattigan]

...
since our integral is between V2 and V1, we do ln|V2| - ln|V1| = ln|V2/V1|

$$\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{. ...}$$

$$\text{ ... Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}$$

(As before.)

 

[jamesd2008]

Thanks everyone for there help.

 

[Char. Limit]

$$\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{.}$$

$$\text{Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}$$

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.