Fundamental theorem of calculus, Taylor's theorem 
I don't really see why this follows? How to demonstrate that this is true? 
I've tried using $f(x)  = x^2$, but that didn't work. 
Why does $f(b) = f(a) + \int_a^b f'(t) dt$ ? 
 A: The Fundamental Theorem of Calculus states that if $f$ is integrable on $[a,b]$ and $F$ is a function satisfying $F'(x)=f(x)$ for all $x \in (a,b),$ then
$$ F(a)-F(b)= \int_{a}^b f(t)dt.$$
In your case they write $F$ as $f$ and $f$ as $f'.$
A: Beacause that's what the Fundamental theorem of Calculus says.
And why do you say that it does not work for $f(x)=x^2$? It does, since $f'(x)=2x$ and$$\int_a^b2t\,\mathrm dt=b^2-a^2=f(b)-f(a).$$
A: Probably, you have seen this theorem written as: $$\int_{a}^{b}f'(x)dx = f(b) - f(a)$$
And now you only has to pass the term $f(a)$ to the left.
A: Let $n>1$ and  $x_k=a+k\frac{b-a}{n} $
then
$$f (b)-f (a)=\sum_{k=1}^n \Bigl(f (x_k)-f (x_{k-1})\Bigr)$$
by MVT,
$$=\sum_{k=1}^n (x_k-x_{k-1})f'(c_k) $$
which is a Riemann sum of the integrable function $f'$.
This sum does not depend on $n $.
when $n\to +\infty $ it converges to the integrale.
hence $$\int_a^bf'(x)dx=f (b)-f (a) $$
A: Short answer: fundamental thm of calculus $\int_a^b f'(t)dt = f(b) - f(a)$ follows because $f(x)$ is an antiderivative of $f'(x)$, i.e. a function $F(x)$ that verifies $F'(x)=f'(x)$ can be just $f(x)$.
Another try: Define $g(x)=f'(x)$. By the fundamental theorem of calculus, if $G'(x)=g(x)$ then $\int_a^b g(t)dt = G(b) - G(a)$. An antiderivative of $g$ is differentiable function $G$ such that $G'(x)=g(x)$, hence we need $G$ such that $G'(x)=f'(x)$. Setting $G(x)=f(x)$ works (because then $f'(x)=f'(x)$). Hence $G'(x)=g(x)$ then $\int_a^b g(t)dt = G(b) - G(a)$ amounts to $\int_a^b f'(t)dt = f(b) - f(a)$.
