Integration proofs help (1 Viewer)

[LOL™]

Hey!! can anyone help me with this? I gave up after like 1/2 hr

If f(x) = f(a - x), prove that

thanks

 

[MOP777]

Pretty sure this is correct. I did skip a few steps but it's all correct as far as I know.

 

[LOL™]

How did you get from the 4th to the 5th line
I dont get how:
<a href="http://www.codecogs.com/eqnedit.php?latex=2\int_{0}^{a}xf(a-x) = I" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2\int_{0}^{a}xf(a-x) = I" title="2\int_{0}^{a}xf(a-x) = I" /></a>
Because that wouldn't be f(a-x)

 

[MOP777]

because f(a-x) = f(x) as originally stated, and as stated in the proof.

 

[LOL™]

But f(a-x) would be
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{a}(a-x)f(a-x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{a}(a-x)f(a-x)" title="\int_{0}^{a}(a-x)f(a-x)" /></a>

rather than this

How did you get from the 4th to the 5th line
<a href="http://www.codecogs.com/eqnedit.php?latex=2\int_{0}^{a}xf(a-x) = I" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2\int_{0}^{a}xf(a-x) = I" title="2\int_{0}^{a}xf(a-x) = I" /></a>

 

[MOP777]

You split the (a-x) function. Because it is (a-x)*(f(a-x)) it equals a*(f(a-x)) - x*(f(a-x))

Do you get that? I'll give you an example without variables. (5-2) *(f(x)) = 5*(f(x)) - 2*(f(x))
Do you understand?

 

[LOL™]

You split the (a-x) function. Because it is (a-x)*(f(a-x)) it equals a*(f(a-x)) - x*(f(a-x))

Do you get that? I'll give you an example without variables. (5-2) *(f(x)) = 5*(f(x)) - 2*(f(x))
Do you understand?

Yea i get that of course but how does


Because its clearly not f(a - x)

 

[MOP777]

Because f(a-x) = f(x)
ie


I don't know how to explain it any clearer.

excuse the

< br / >

I can't get rid of them

 

[LOL™]

Because f(a-x) = f(x)
ie


I don't know how to explain it any clearer.

excuse the

< br / >

I can't get rid of them

lol
the point i don't get is how that is = I
because the function there lets call it g(x) is xf(x)
therefore g(a-x) would have to be (a-x)f(a-x) and not xf(a-x)
Do you get what im trying to ask ? lol sorry

 

[MOP777]

I do get what you're saying
but (a-x) TIMES f(a-x)
= (a TIMES f(a-x)) MINUS (x TIMES f(a-x))

you split the a minus x into two separate integrals.

 

[Trebla]

Would this approach help?

 

[seanieg89]

You've created an extra factor of x in the integrand of the RHS of your final line trebla.

 

[jet]

lol
the point i don't get is how that is = I
because the function there lets call it g(x) is xf(x)
therefore g(a-x) would have to be (a-x)f(a-x) and not xf(a-x)
Do you get what im trying to ask ? lol sorry

Becuase when they refer to f(x) they aren't talking about the integral, they're just talking about the function inside the integrand. So, since f(a -x) = f(x) then


The fact that the first poster replaced every instance of x by a-x at the start of his proof uses a completely different rule altogether.

 

[LOL™]

Becuase when they refer to f(x) they aren't talking about the integral, they're just talking about the function inside the integrand. So, since f(a -x) = f(x) then


The fact that the first poster replaced every instance of x by a-x at the start of his proof uses a completely different rule altogether.

ah ok i think i get what you're trying to say now. So in the question when its says given that f(x) = f(a-x), it is actually saying that those 2 functions are equal rather than refering to the general rule.
So then in the function xf(a-x), you can simply replace the f(a-x) by f(x) so then it would become xf(x)? right?

anyway, thanks for that everybody.

 

[jet]

Yes that's right