# I Integration of 1 variable in 2 different ways.

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1. Apr 14, 2016

### sreerajt

I have to do a integration which goes like this:
(V-M)(dP/dx)+3P(dV/dx)=0, (where M,P and V are constants).
If you integrate with dx, you will get:
∫[(V-M)dP]+∫[3PdV]=0.
which ultimately results in the answer M=4V.
Now, i can put the first equation in this form also:
(dP/P)=-3[dV/(V-M)]. Integration will give you,
lnP=-3ln(V-M) , where 'ln' is the logarithm to base e. This give lnP+ln[(V-M)3]=0.
This gives entirely different answer. So which is the correct way??

2. Apr 14, 2016

### Ssnow

In both you forget the constant integration $c$ ...

3. Apr 14, 2016

### Samy_A

Something doesn't add up here.
If $V, P$ are constants, then $\frac{dV}{dx}=0, \frac{dP}{dx}=0$. Your equation then becomes $0+0=0$.

4. Apr 14, 2016

### Ssnow

Second as @Samy_A said not all can be constant ...

5. Apr 14, 2016

### sreerajt

OMG . Ya, that's correct.
But, if P and V is a function of x, then ??? Infact i did that calculation keeping this in mind that P and V is a function of x.

6. Apr 14, 2016

### Samy_A

If V and P are functions, how can you integrate ∫[(V-M)dP]+∫[3PdV] the way you did?
Only your second method makes sense, and as @Ssnow mentioned, don't forget the integration constant.

7. Apr 14, 2016

### sreerajt

I made lot many mistakes....
thanks for pointing out...
thank you...