Why is this continuous representation differentiable? Let $V$ be a real or complex finite-dimensional vector space and $\pi$ a continuous representation of the additive group $\mathbb{R}$ on $V$:
$$
\pi (t+s) = \pi (t) \pi (s), \ t,s\in\mathbb{R}, \ \pi(0)=I.
$$
Prove that $\pi : \mathbb{R} \to L(V)$ is differentiable.
I have a hint:
Note that from the continuity of $\pi$ we have
$$
\lim_{t\to 0} \frac{1}{t} \int_0^t \! \pi(t) \,\mathrm{d} t = \pi(0) = I.
$$
I don't see how this is obvious. What troubles me is the fact that there is an integral sign in the above equation, where did it come from? And because $\pi(t)$ is a linear map, how do I integrate it?
 A: The integral sign comes by fiat. The fundamental theorem of calculus tells us that the integral of a continuous function is (continuously) differentiable, introducing an integration gives us a more regular function to work with. While we only know that $\pi$ is continuous, we know that the integral of $\pi$ is differentiable, and therefore we introduce the integral. That is later used to deduce the differentiability of $\pi$.

And because $\pi(t)$ is a linear map, how do I integrate it?

Choose a basis $B$ of $V$, and identify $\pi(t)$ with its matrix representation with respect to $B$. Then take the componentwise integral, the entry in the $i^{\text{th}}$ row and $j^{\text{th}}$ column of $\int_0^t \pi(s)\,ds$ is the integral of the corresponding component function.
We can define the integral of $L(V)$-valued functions more abstractly, and when $V$ is infinite-dimensional, that is necessary, but in finite dimensions, it is easier to just use a matrix representation.
Finally, for continuous real- or complex-valued functions $f$ the fact that
$$\lim_{t\to 0} \frac{1}{t}\int_0^t f(s)\,ds = f(0)$$
is probably familiar. And since that holds for all components of the matrix function, it holds for the entire matrix function.
