Let $u: [a,b] \to X$ to be a continuous function and $X$ a Banach space. Then $$ \frac{1}{b - a} \int_{a}^{b} u(t) \ \text{d}t \in \overline{\text{co}}(u([a,b])) $$ holds, where $\overline{\text{co}}(M) = \overline{\text{co}(M)}^{\| \cdot \|_{X}}$ is the closed convex hull for some set $M \subset X$ and $$ \text{co}(M) := \left\{ \sum_{i = 1}^{N} \lambda_i x_i: N \in \mathbb{N}, \lambda_i \in [0,1], \sum_{i = 1}^{N} \lambda_i = 1, x_i \in M \ \forall i \in \{1, \ldots, N\} \right\}. $$
How can we visualise this result geometrically?
The integral on left-hand side (divided by $b-a$), which I denote by $\bar u$ in the visualization, can be interpreted as an average over the trajectory of $u$. I have tried to visualize it for two different trajectories in $\mathbb R^2$. In some sense, it is a weighted average over the points of the trajectory, since "segments" where the velocity is high get a lower weight in the average process. I tried to visualize this fact by the intermediate points, in particular in the second example, where the velocity is high in the beginning and low in the end, thereby shifting $\bar u$ to the right. The true average would be the mean of the images of infinitely many equidistant "intermediate points" in $[a,b]$ (or, alternatively, one may think of the segments between those points).
As mentioned before, $\bar u$ being a (weighted) average over the trajectory guarantees that it lies in the closed convex hull of the trajectory