Fundamental theorem of calculus proof? To give you background, I have been studying calculus from Stewart. In the Integrals chapter, he proved the Evaluation Therorem by applying the mean value theorem on the Riemann sum of a continous function $f$:
$$\int_a^bf(x)dx = F(b)−F(a).$$ 
Now when I studying the Fundamental Theorem of Calculus, I think that it is just a corollary of the theorem above. Then why does he go through the proof using some other method? It is really something that is bothering me.
 A: What Stewart, but no one else, calls the Evaluation theorem:
$$\int_a^b f(x) \,dx = F(b) - F(a), \qquad \text{where } F' = f$$
is actually true in the slightly greater generality where we only assume that $f$ is Riemann integrable, but not necessarily continuous.  On the other hand, what you presumably consider the Fundamental Theorem of Calculus (which the above is also usually called):
$$ \frac{d}{dx} \int_a^x f(t) \, dt = f(x)$$
is not true in this generality, but only for continuous functions $f$.  Now, the first theorem is a corollary of the second when $f$ is continuous, but when it is not, a separate proof is needed.  It may appear that the second theorem is a corollary of the first under any conditions (just differentiate both sides with respect to $b$), but there is a catch: the "Evaluation Theorem" assumes the existence of an antiderivative $F$, while the "Fundamental Theorem" constructs that antiderivative.  So the Evaluation Theorem can never prove the Fundamental Theorem without being circular or redundant.
If Stewart states continuity as a hypothesis, then the proof of the Evaluation Theorem alone is simply redundant.  This is not to say it is pointless, though, since it provides an entirely different explanation of the phenomenon of "differentiation and integration are inverses", and hints at the importance of the mean value theorem in using a function's derivative to affect the function itself.
