Integral proof and then use the proof to solve a 2nd integral

[Mustard]

Homework Statement

I'm not sure how to go about it tbh :/

Relevant Equations

On the pic

 

[PeroK]

Do you have any ideas yourself?

 

[Mustard]

Do you have any ideas yourself?

Well in the the bottom half I would assume you would have to use u-subsitution but I don't believe the question is asking me to solve the bottom until I prove the top is = b/2?

 

[PeroK]

Well in the the bottom half I would assume you would have to use u-subsitution but I don't believe the question is asking me to solve the bottom until I prove the top is = b/2?

Have you tried the obvious substitution?

 

[Mustard]

Have you tried the obvious substitution?

Do you mean substituting b for x?

 

[PeroK]

Do you mean substituting b for x?

I thought the substitution $u = b - x$ was the first thing you should consider. Especially if you are stuck.

 

[Mustard]

I thought the substitution $u = b - x$ was the first thing you should consider. Especially if you are stuck.

Oh , I'm stuck again. Am I doing it right si far ? :/ I'm sorry it's just confusing to me.

 

[PeroK]

Oh , I'm stuck again. Am I doing it right si far ? :/ I'm sorry it's just confusing to me.

Yes. You need to tidy that up and, strictly speaking, you are missing an equals sign.

 

[SammyS]

Oh , I'm stuck again. Am I doing it right si far ? :/ I'm sorry it's just confusing to me.

You have a few issues to correct.

If $u=b-x$, then $x=b-u$ not $u-b$.

The denominator is a sum, $f(x)+f(b-x)$, not a difference.

The integration limits for $u$ are different than for $x$.

After making these corrections, there will be more work to do to complete your task.

 

[Delta2]

After you do the necessary corrections outlined in post #9 , you ll have the integral $$\int_b^0-\frac{f(b-u)}{f(b-u)+f(u)}du$$, which is the same as $$\int_0^b \frac{f(b-u)}{f(b-u)+f(u)}du$$. Now you ll have to do a little algebraic trick (adding and subtracting from the numerator the same quantity f(u) )and then by algebraic simplifications you ll get a significant result, namely that the initial integral (the integral involving x, let's call it I) is such that $$I=b-I$$. And this is pretty much the end of it, solving the last equation for I.

 

[epenguin]

Looks to me an exercise in My First Math Book, Chapter 1, Symbols and Functions.

Ex 15: If φ$(f,b) =$ ψ$(b)$ what is φ$(f, 3)$?

if you can calculate ψ(3) what can you say about φ$(f,3)$ ?