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kalagota · #2 November 10th, 2007, 10:03 PM

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

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

Let $f_0(x) = x^n f(x)$ and $g_0(x) = x^n g(x)$

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)

$\implies x^n f(x) = x^n g(x)$

$\implies f(x) = g(x)$ . qed