information:classification
Relationship between the parts

As discussed above, a slightly weaker version of the second part follows from the first part.

Similarly, it almost looks like the first part of the theorem follows directly from the second. That is, suppose G is an antiderivative of f. Then by the second theorem, G (x) − G (a) = ∫ a x f (t) d t {\textstyle G(x)-G(a)=\int _{a}^{x}f(t)\,dt}. Now, suppose F (x) = ∫ a x f (t) d t = G (x) − G (a) {\textstyle F(x)=\int _{a}^{x}f(t)\,dt=G(x)-G(a)}. Then F has the same derivative as G, and therefore F ′ = f. This argument only works, however, if we already know that f has an antiderivative, and the only way we know that all continuous functions have antiderivatives is by the first part of the Fundamental Theorem. For example, if f (x) = e, then f has an antiderivative, namely G (x) = ∫ 0 x f (t) d t {\displaystyle G(x)=\int _{0}^{x}f(t)\,dt} and there is no simpler expression for this function. It is therefore important not to interpret the second part of the theorem as the definition of the integral. Indeed, there are many functions that are integrable but lack elementary antiderivatives, and discontinuous functions can be integrable but lack any antiderivatives at all. Conversely, many functions that have antiderivatives are not Riemann integrable (see Volterra's function).