Given a function $\displaystyle f(x)$ defined on interval $\displaystyle [a,b]$.Provethat there must exist a function $\displaystyle F(x)$ with the property that $\displaystyle F'(x)=f(x)$ $\displaystyle \forall x\in (a,b)$.

With this we can prove the second fundamental theorem of calculus. We have to show that if $\displaystyle g(x)=\int^x_a f(x)dx$ then $\displaystyle g'(x)=f(x)$. Instead of the classical proof with a Riemann Sum we may do the following: Since by the first fundamental theorem of calculus, "If there exists an anti-derivative of $\displaystyle f(x)$ then $\displaystyle \int^x_a f(x)dx = F(x)-F(a)$" But by the existence of anti-derivative conjecture there MUST exist such a function thus, $\displaystyle g(x)=F(x)-F(a)$ but then $\displaystyle g'(x)=f(x)$ because $\displaystyle F(a)$ is a constant-function.