 Quote by mishin050
$$ \frac{d}{dx} \int_a^x f(t)~dt = f(x);$$
$$ dx \cdot\frac{d}{dx} \int_a^x f(t)~dt = f(x)\cdot dx;$$
$$\int d \int_a^x f(t)~dt =\int f(x)~dx;$$
$$ \int_a^x f(t)~dt =\int f(x)~dx;$$
$$\int f(x)~dx= F(x)-F(a)!$$
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In your own (incorrect anyway) proof, you used the fact that [itex]\int f(x) dx=F(x)[/itex] between the third and fourth lines. You weren't correct getting to that line, but even if you were, you assume the exact fact that you are trying to prove wrong. Here's my proof...
Let the operator [itex]\int dx[/itex] denote the operation that sends a function [itex]f[/itex] to the equivalence class of antiderivatives [itex]F+C[/itex] where [itex]F[/itex] is any antiderivative and [itex]C[/itex] can be any element of the range of the function F.
Therefore,
$$\int f(x)dx=F(x)$$
because it is the definition! There's no math involved. The fundamental theorem makes this notation meaningful, but this is merely notation that was chosen because it makes sense.
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