Inequality involving norm of matrix integral

This question seems basic but I could not find an answer. I have seen the inequality $$\left\|\int_a^b x(t) dt \right\| \leq \int_a^b \left\| x(t) \right\| dt$$ where $x(t) \in \mathbb{R}^n$ is a vector function and $\|\cdot\|$ is a vector norm, and $a < b$.

I wonder if this also holds for matrices with induced norm, that is $$\left\|\int_a^b X(t) dt \right\| \leq \int_a^b \left\| X(t) \right\| dt$$ where $X(t)$ is a matrix function and $\|\cdot\|$ is an induced matrix norm, and $a < b$. If it is true, is there any reliable citation source?

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By induced norm, you mean a norm of the form $\sup_{v\neq 0}\frac{||Av||}{||v||}$? –  Davide Giraudo Feb 29 '12 at 20:41
The space $M_n(\mathbb{R})$ is just a finite dimensional vector space like any other, so you can directly apply the the triangle inequality to any matrix-valued function, with any norm, induced or not (actually you could even have non square matrices). Induced norm are useful when dealing with matrix product but there's none here. –  Joel Cohen Mar 1 '12 at 0:51

If Riemann-integrals are good enough for you, then these inequalities are just the disguised triangle inequality (here with sloppy notation): $$\left\|\int_a^b X(t) dt \right\| = \left\| \lim_{\mathcal Z} \sum_{\mathcal Z}X(\xi) \right\| \leq \lim_{\mathcal Z} \sum_{\mathcal Z} \left\| X(\xi) \right\| \leq \int_a^b \left\| X(t) \right\| dt$$
If $||\cdot||$ is given by $||A||=\sup_{v\neq 0}\frac{N(Av)}{N(v)}$ then we have for a fixed $v$ $$N\left(\int_a^bX(t)dtv\right)=N\left(\int_a^bX(t)vdt\right)\leq \int_a^bN\left(X(t)v\right)dt\leq \int_a^b||X(t)||N(v)dt$$ so $\|\int_a^b X(t)dt\|\leq \int_a^b||X(t)||dt$.
In fact more generally, if $f\colon [a,b]\to X$ where $(X,||\dot||$ is a Banach space then $\|\int_a^b f(t)dt\|\leq \int_a^b||f(t)||dt$. It can be showed using a corollary of Hahn-Banach theorem: \begin{align*}\|\int_a^b f(t)dt\|&=\sup_{\varphi\in X',||\varphi||=1}\left|\varphi\left(\int_a^b f(t)dt\right)\right|\\ &=\sup_{\varphi\in X',||\varphi||=1}\left|\int_a^b \varphi\left(f(t)\right)dt\right|\\ &\leq \sup_{\varphi\in X',||\varphi||=1}\int_a^b \left|\varphi\left(f(t)\right)\right|dt\\ &\leq \sup_{\varphi\in X',||\varphi||=1}\int_a^b \|f(t)\|dt \\ &=\int_a^b \|f(t)\|dt. \end{align*}