I just thought of the following question, which at least holds for $1$, $2$ and ${\infty}$ norm in $\mathbb {R}^n$.
Let $X= \mathcal{C}([a,b], \mathbb{R}^n)$ be the set of bounded continuous functions from $[a,b]$ to $\mathbb{R}^n$. Then for all $g \in X$, and any norm $|| \cdot || $ of $\mathbb{R}^n$ we have $$ \bigg|\bigg| \int_{a}^b g(s) ds \bigg|\bigg| \le \int_{a}^{b} || g(s) || ds $$ where integral of an vector in $\mathbb{R}^n$ is defined component wise.
I hope I had not make any silly generalization. Is this true?
The Riemann sum of $g$ with respect to the partition $\mathcal{P}_n=\{x_k=a+k\frac{b-a}{n}\}_{k=0}^n$ satisfies your claim by the triangular inequality property of norms; $$ \lvert \lvert \sum_{k=1}^ng(x_k^*)(x_k-x_{k-1})\rvert\rvert\leq\sum_{k=1}^n||g(x_k^*)||(x_k-x_{k-1}),\hspace{.7cm}x_{k-1}\leq x_k^*\leq x_k. $$ As $g$ is integrable (for it is a continuous function defined in a compact set) and as the norm is a continuous function we can take the limit as $n\to\infty$ to conclude $$ \lvert\lvert\int_a^bg(s)ds\rvert\rvert\leq\int_a^b||g(s)||ds. $$