What's the derivative of $\int_0^x e^{t^2} dt$? Let f be a continuous function on the interval $[a, b]$. The function F defined by  
$$ \mathcal F(x) = \int_a^x f(t)dt $$ 
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$\mathcal F'(x) = \mathcal f(x)$$
My question is the following: What will happen in this case?
$$\mathcal H(x) = \int_0^x e^{t^2} dt$$
Would the derivative be: 
$$\mathcal H'(x) = \mathcal e^{x^2}$$
or
$$\mathcal H'(x) = \mathcal e^{x^2}-1$$
 A: Note that $F(x+h)= F(x) + \int_x^{x+h} f(t)dt$ and for small $h$ we have $\int_x^{x+h} f(t)dt \approx \int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.
It is straightforward to make this argument rigorous.
A: If we use the idea in your comment (to the question) then $\mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).
Part 1 of FTC deals with an integral of the form $\int_{a} ^{x} f(t) \, dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $\mathcal{F} :[a, b] \to\mathbb{R} $ via the relation $$\mathcal {F} (x) =\int_{a} ^{x} f(t) \, dt\tag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $\mathcal{F} $ in terms of properties of $f$. And it says that $\mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $c\in[a, b] $ then $\mathcal{F} $ is differentiable at $c$ and $\mathcal{F} '(c) =f(c) $.
You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $\mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $\mathcal{F}' (x) $ depends on $f$ and $x$ only.
Part 2 of FTC deals with the evaluation of $\int_{a} ^{b} f(x) \, dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $\mathcal{F} $ so that $$\mathcal{F} '(x) =f(x), \forall x\in[a, b] $$ and then says that $$\int_{a} ^{b} f(x) \, dx=\mathcal{F} (b) - \mathcal {F} (a) \tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.
Note that the $\mathcal{F} $ in both parts of FTC are different and in particular the $\mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.
A: Recall that in general by Leibniz integral rule the following holds
$$F(x)=\int_{a(x)}^{b(x)}g(u) du\implies F'(x)=g(b(x))\cdot b'(x)-g(a(x))\cdot a'(x)$$
therefore
$$\mathcal H(x) = \int_0^x e^{t^2} dt\implies \mathcal H'(x)=e^{x^2}$$
A: The simplest when applying a new formula is to identify each component:

Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$F(x) = \int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$F'(x) = \mathcal f(x)$$

For $H(x) = \int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.
It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $\mathbb R$, so their composition $f=g\circ h$ is continuous on $\mathbb R$, hence on $[a,b]$.
Now, we can safely apply the formula: $H'(x) = f(x)=\mathcal e^{x^2}$.

Edit: to answer a comment

What would happen if $a$ is not $0$?

Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=\int_1^{x} f(t)\; dt= \int_1^x e^{t^2}\; dt$. Then, all of the above applies here and we have
$$H_2'(x) = f(x) = e^{x^2} $$
Wait! Why do the functions $H_2$ and $H_1(x)=\int^x_0 f(t)\; dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that
$$H_1(x)=\int_0^x f(t)\; dt=\int_0^1 f(t)\; dt + \int_1^x f(t)\; dt= C + H_2(x)$$
where $C=\int_0^1 f(t)\; dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.
