Integration by parts involving an unknown function

In summary, the conversation is about a question regarding the use of integration by parts to solve an integral. The person is confused about how the equations (2.13) and (2.14) are related and does not understand why (2.14) does not depend on sigma(t). The other person suggests rewriting one of the terms as an integral to better understand the relationship between the two equations.
  • #1
sara_87
763
0

Homework Statement



I have attached a picture including 2 equations: (2.13) and (2.14)
I don't understand how they got from (2.13) to (2.14) using integration by parts

Homework Equations





The Attempt at a Solution


For the integral:
[itex]\int_{\tau_0}^t\sigma(\tau)d\tau= \left[\sigma(\tau)\tau\right]_{\tau_0}^{t}-\int_{\tau_0}^t\dot{\sigma}(\tau)\tau d\tau[/itex]
by assuming [itex]u=\sigma(\tau)[/itex] and [itex]v'=1[/itex] for integration by parts.
This gives:
[itex]\sigma(t)t-\sigma(\tau_0)\tau_0-\int_{\tau_0}^{t}\tau\dot{\sigma}(\tau)d\tau[/itex]

So, even when we do combine this result with the other terms in equation (2.13), I don't understand how (2.14) does not depend on [itex]\sigma(t)[/itex].

Thank you in advance for any ideas.
 

Attachments

  • ibp.jpg
    ibp.jpg
    11.9 KB · Views: 882
Physics news on Phys.org
  • #2
Hi sara_87! :)

One of your terms is ##\sigma(t)t##.
You can rewrite that as an integral:

$$t \sigma(t)+C=t \int_{\tau_0}^t \dot\sigma(\tau)d\tau=\int_{\tau_0}^t t \dot\sigma(\tau)d\tau$$

The variable t can be moved into the integral, since you do not integrate over t, so t behaves like a constant with respect to integration.
 

What is integration by parts involving an unknown function?

Integration by parts is a method used in calculus to find the integral of a product of two functions. When one of the functions is unknown, integration by parts involves using the product rule in reverse to solve for that unknown function.

Why is integration by parts involving an unknown function useful?

This method is useful when the integral of a function cannot be easily found by other methods. It allows for the integration of more complex functions by breaking them down into simpler parts.

What is the general formula for integration by parts involving an unknown function?

The formula is given by ∫uv' dx = uv - ∫u'v dx, where u and v are the two functions and u' and v' are their derivatives.

How do you choose which function to assign as u and v in integration by parts involving an unknown function?

The acronym "LIATE" can be helpful in choosing u and v. L stands for logarithmic functions, I for inverse trigonometric functions, A for algebraic functions, T for trigonometric functions, and E for exponential functions. The function assigned as u should be the first in the acronym that appears in the integral.

Can integration by parts be used for definite integrals involving unknown functions?

Yes, integration by parts can be used for both indefinite and definite integrals involving unknown functions. In the case of definite integrals, the limits of integration should be substituted into the final equation after solving for the unknown function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
646
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
12
Views
1K
  • Differential Equations
Replies
1
Views
676
Replies
4
Views
609
  • Calculus and Beyond Homework Help
Replies
1
Views
947
  • Atomic and Condensed Matter
Replies
1
Views
813
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
26K
Back
Top