Integration of 1 variable in 2 different ways.

In summary, if V and P are functions, and you integrate ∫[(V-M)dP]+∫[3PdV] the way you did, you will get two different answers depending on whether P and V are functions of x or not.
  • #1
sreerajt
39
1
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??
Thanks in advance...
 
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  • #2
In both you forget the constant integration ##c## ...
 
  • #3
sreerajt said:
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).
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##.
sreerajt said:
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??
Thanks in advance...
 
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  • #4
Second as @Samy_A said not all can be constant ...
 
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  • #5
OMG :nb). Ya, that's correct.
But, if P and V is a function of x, then ? In fact i did that calculation keeping this in mind that P and V is a function of x.
 
  • #6
sreerajt said:
OMG :nb). Ya, that's correct.
But, if P and V is a function of x, then ? In fact i did that calculation keeping this in mind that P and V is a function of x.
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.
 
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  • #7
Samy_A said:
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.
I made lot many mistakes...
thanks for pointing out...
thank you...
 
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Related to Integration of 1 variable in 2 different ways.

1. What is integration of 1 variable in 2 different ways?

Integration is a mathematical process that involves finding the area under a curve. When we talk about integration of 1 variable in 2 different ways, it means that we are using two different mathematical techniques to find the area under the same curve.

2. Why would we need to use 2 different ways to integrate 1 variable?

Using 2 different ways to integrate 1 variable can help us check our answers and verify that they are correct. It also allows us to approach the problem from different perspectives and gain a deeper understanding of the mathematical concepts involved.

3. What are the two most commonly used methods for integration of 1 variable?

The two most commonly used methods for integration of 1 variable are the Riemann sum method and the Fundamental Theorem of Calculus. The Riemann sum method involves dividing the area under the curve into smaller rectangles and calculating their individual areas, while the Fundamental Theorem of Calculus uses derivatives and antiderivatives to find the area.

4. Can both methods give different results for the same problem?

Yes, it is possible for the Riemann sum method and the Fundamental Theorem of Calculus to give different results for the same integration problem. This can happen when the function being integrated is not continuous or has sharp turns, making it difficult to accurately approximate the area using rectangles.

5. Are there any other methods for integration of 1 variable?

Yes, there are other methods for integration of 1 variable, such as the Trapezoidal Rule, Simpson's Rule, and Monte Carlo integration. These methods have their own advantages and are often used in different fields of science and engineering.

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