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ThePerfectHacker · #3 November 10th, 2007, 10:16 PM

I am sorry, I cannot follow.

Quote:

Originally Posted by kalagota

f and g are cont on [0,1], then f and g are integrable.

Let $F_0 = \int_0^1 x^n f(x)dx = \int_0^1 x^n g(x)dx$

How is this a well-defined function? Maybe for a particular value of $n$ ?

Quote:

then both $f_0, g_0$ are continuous on [0,1], which implies that $F_0$ is differentiable on [0,1], and $F_0'(x) = f_0(x) = g_0(x)$ , for all $x \in [0,1]$ (by a corollary to the FTOC)

That is not the Fundamental Theorem of Calculus.
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Remember I am saying that:
$\int_0^1 f(x)dx = \int_0^1 g(x) dx$ , $\int_0^1 xf(x) dx = \int_0^1 xg(x) dx$ , $\int_0^1 x^2f(x) dx = \int_0^1 x^2g(x) dx$ , ...