Integration by parts with absolute function

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Discussion Overview

The discussion revolves around the integration by parts involving an absolute function, specifically focusing on calculating the average value of a function between specified limits. Participants are exploring methods to solve for a variable denoted as P, with some expressing difficulties due to the absolute value of the cosine function.

Discussion Character

  • Technical explanation, Homework-related, Mathematical reasoning

Main Points Raised

  • One participant, Mathsboi, presents a problem involving the average value of a function and expresses difficulty in solving it due to the absolute value of the cosine function.
  • Another participant provides a numerical approximation for P using a computer algebra system (CAS), stating that $$p\approx\frac{17812}{13715}$$.
  • A follow-up request for clarification on the method used to obtain the value of P is made, along with a query about finding P with different left-hand side values.
  • Another participant suggests factoring P out of the integrand as a potential approach to simplify the problem.

Areas of Agreement / Disagreement

The discussion includes multiple viewpoints and approaches to solving the problem, with no consensus reached on a single method or solution.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the integration, and there are dependencies on specific values and definitions that remain unclear.

Mathboi1
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Hi all,

I have the average value of a function between limits of 7.3826 and 0 which equals 0.4453. I have used the formula for average value function and attached the equation I need solving as I don't know how to use the Latex commands. P is what I am trying to work out. Unfortunately I have been unsuccessful since the absolute of cos makes this difficult. If anyone can solve this with working solution I'd be very grateful as I've spent all day trying to solve it.

View attachment 5087

Cheers,

Mathsboi
 

Attachments

  • power equation.png
    power equation.png
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Using a CAS, I get:

$$p\approx\frac{17812}{13715}$$
 
MarkFL said:
Using a CAS, I get:

$$p\approx\frac{17812}{13715}$$

Thanks for responding,

Can you please explain how you got here? I need to find p with different values of the LHS value and 0.125, keeping everything else the same. Is there a general formula I could use or can you show me steps to your solution?
 
Last edited:
Factor $p$ out of the integrand.
 

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