Solution for Tricky Definite Integral: How to Find I in Terms of A"

[jam_27]

Am trying to get a solution to the definite integral below. Looking for some direction.

I = ₀∫¹ xf(x)dx where

₀∫¹ f(x)dx = A, is known.

Also, its is know that when x =1, f(x) =0 and when x =0, f(x) =1.

Can we get a solution of I in terms of A?

I have tried going the integration by-parts route which did not lead to any success. Any help is much appreciated.

 

[mfb]

Without more details about f(x), this is impossible. It is easy to find some examples where the constraints are met, A is the same but I differs.

 

[jam_27]

Unfortunately, f (x) is a transcendental function

f(x) = a - be^[px+qf(x)] -c[px+qf(x)]

where, a, b, c, p and q are all constants. I don't think this helps?

 

[mfb]

I don't know if that helps, but it is at least some hope to make it possible (I just don't know how). Without knowing anything about f(x) it would be completely impossible.

 

[micromass]

It's going to be annoying, but if $f$ has an expansion as a power series, then you might be able to find an approximation to $I$. I really doubt a analytic solution exists.

 

[micromass]

Wolfram alpha gives the following (very ugly) form of $f(x)$: http://www.wolframalpha.com/input/?i=x+=+a+-+b*e^(p*r+q*x)+-c*(p*r+q*x)

(just replace $x$ by $f(x)$ and $r$ by $x$). So yes, it's not going to be pretty.

 

[jam_27]

Wolfram alpha gives the following (very ugly) form of $f(x)$: http://www.wolframalpha.com/input/?i=x+=+a+-+b*e^(p*r+q*x)+-c*(p*r+q*x)

(just replace $x$ by $f(x)$ and $r$ by $x$). So yes, it's not going to be pretty.

wolframalpha gives the result in terms of Lambert's W function which I have already looked at. Looking for another route...

 

[jam_27]

I am wondering if I can use rotation of coordinates to solve this integral, like here, Example D.9? Looking for some direction...

 

[csleong]

try this way, set u=x, dv=f(x)dx, you get du=dx and v=∫f(x)dx
then I=Ax-∫∫f(x)dxdx. Can you solve this?

 

[jam_27]

try this way,...Can you solve this?

Nope, not possible with what I know.