How Does Integration Apply to DW/RDR in Calculus?

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SUMMARY

The discussion focuses on integrating the expression dw/dr with respect to r, specifically the integral \int \frac{1}{r} \frac{dw}{dr} dr. Participants clarify that the integration process depends on the specific function w(r), which is currently unknown. The integration by parts method was attempted but yielded a zero result, indicating a misunderstanding of the function's nature. Ultimately, the fundamental theorem of calculus provides the integral solution as w(r) + C, where C is an arbitrary constant.

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harpreet singh
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Plz explain me how can i integrate dw/rdr
 
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please help me with this..
 
harpreet singh said:
Plz explain me how can i integrate dw/rdr
Could you perhaps explain you question further?
 
w is a function of r and i need to integrate dw/rdr with respect to r
 
harpreet singh said:
w is a function of r and i need to integrate dw/rdr with respect to r

you mean you want to find

\int \frac{1}{r} \frac{dw}{dr} dr


?
 
Ya exactly..
 
I was trying it by using by parts.. By taking 1/r as the first function. But with that i m getting the answer zero..
 
harpreet singh said:
I was trying it by using by parts.. By taking 1/r as the first function. But with that i m getting the answer zero..

Do you happen to know what dw/dr is ?
 
w is a function of r here.. So it can be written as dw(r)/dr
 
  • #10
I'm sorry but this just seems silly. It's like asking, how do I do the integral \int{f(x)dx}. The answer to this question will clearly depend on what f(x) is. Here, the answer to your question will depend on what kind of function w(r) is. So unless you give us more information, I don't think we can help you.
 
  • #11
Actually I am calculating the value of w(r) from a differential equation.. And in satisfying the boundary conditions I need to know that integral to calculate the valuues of constants. As of now w(r) is unknown.
 
  • #12
Then I suggest you post the question. We won't know how to help if we don't know exactly what the question is asking for.
 
  • #13
Going back to the original question, which I suspect is NOT the question you are really asking,

\int (dw/dr)dr= w(r)+ C<br /> where C is an arbitrary constant, by the fundamental theorem of calculus.
 

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