# Tricky definite integral

1. May 4, 2014

### jam_27

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

I = 01 xf(x)dx where

01 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.

2. May 4, 2014

### Staff: Mentor

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.

3. May 4, 2014

### 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?

4. May 4, 2014

### Staff: Mentor

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.

5. May 4, 2014

### 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.

6. May 4, 2014

7. May 4, 2014

### jam_27

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

8. May 15, 2014

### 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.....

9. May 16, 2014

### 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?

10. May 19, 2014

### jam_27

Nope, not possible with what I know.