Integration by parts of a differential

In summary, the conversation discusses the integration by parts of a complex integral and the confusion around calculating the necessary variables. Suggestions are made to split the integral into two parts and to assign different variables for u and dv in order to find the correct result.
  • #1
maistral
240
17
I'll cut the long story short. What on Earth happened here:

integration.png


I seem to be unable to do the integration by parts of the first term. I end up with a lot of dx's.
 
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  • #2
Try splitting the integral into two parts, with this as the first one:
##\int_{x_i}^{x_j} \frac d{dx}\left( kA\frac{dT}{dx}\right)N_i(x)dx##
 
  • #3
I do know how to split the integral. I just wonder what happened to the identity. Judging from the result, as in the format int(u dv) = uv - int(v du);
u = d/dx (kA dT/dx)
dv = Ni(x)dx

The problem maybe is that I don't know how to calculate du. What would be the result?
 
  • #4
maistral said:
I do know how to split the integral. I just wonder what happened to the identity. Judging from the result, as in the format int(u dv) = uv - int(v du);
u = d/dx (kA dT/dx)
dv = Ni(x)dx

The problem maybe is that I don't know how to calculate du. What would be the result?
Because we're working with nice single-variable functions (presumably), [tex]dq=\frac{dq}{dx} \, dx=\frac{d}{dx}(q) \, dx[/tex] To me, it looks like they let [tex]dv=\frac{d}{dx}\left(kA \frac{dT}{dx}\right) \, dx[/tex] and [tex]u=N_i(x)[/tex] This would imply [tex]v=kA \frac{dT}{dx}[/tex] and [tex]du=dN_i=\frac{dN_i}{dx} \, dx[/tex] which appears to be consistent with the result they got.
 
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  • #5
maistral said:
I do know how to split the integral. I just wonder what happened to the identity. Judging from the result, as in the format int(u dv) = uv - int(v du);
u = d/dx (kA dT/dx)
dv = Ni(x)dx

The problem maybe is that I don't know how to calculate du. What would be the result?
What happens if you instead let ##u = N_i(x)## and ##dv = \frac d {dx} \left(kA \frac{dT}{dx}\right) dx##? Finding v shouldn't be that difficult.
 
  • #6
Thanks!
 

Related to Integration by parts of a differential

1. What is integration by parts?

Integration by parts is a method used in calculus to evaluate integrals of the form ∫u(x)v'(x)dx. It is based on the product rule for differentiation, and allows for the integration of products of functions that cannot be integrated directly.

2. How do I know when to use integration by parts?

Integration by parts is typically used when the integrand is a product of two functions, and one of the functions has a derivative that is easier to integrate than the other. This method is also useful when the integral involves logarithmic or trigonometric functions.

3. What is the formula for integration by parts?

The formula for integration by parts is given by: ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)u'(x)dx. This formula can also be written as ∫u(x)dv(x) = u(x)v(x) - ∫v(x)du(x), depending on the notation used.

4. Can integration by parts be used multiple times?

Yes, integration by parts can be used multiple times in a single integral. This is known as repeated integration by parts. However, each time it is used, the integral will typically become more complicated.

5. What are some common mistakes to avoid when using integration by parts?

Some common mistakes to avoid when using integration by parts include forgetting to apply the formula correctly, choosing the wrong functions for u and v, and not simplifying the integral after each step. It is also important to pay attention to any patterns or simplifications that may arise during the process.

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