The 2-norm of the integral vs the integral of the 2-norm

It's very simple. hehe... (Note you actually want to assume $A,B\in L^1(\Bbb R)$, not $L^2$.)

Edit: Morally the same argument works for a Banach-space valued function; see Below.

To make things easier to type I'm going to revise the notation. Suppose that $f:\Bbb R\to\Bbb R^2$; we want to show that $$\left|\left|\int f(x)\,dx\right|\right|_2\le\int||f(x)||_2\,dx.$$Let $$v=\int f(x)\,dx\in\Bbb R^2.$$Then $$||v||_2^2=v\cdot v=v\cdot\int f(x)\,dx=\int v\cdot f(x)\,dx\le\int||v||_2||f(x)||_2\,dx=||v||_2\int||f(x)||_2\,dx.$$Divide by $||v||_2$: $$||v||_2\le\int||f(x)||_2\,dx.$$

Below One can give a similar argument if $f:S\to X$ where $X$ is a Banach space and $\mu$ is a measure on $S$. Let $$v=\int_Sf(t)\,d\mu(t)\in X.$$Suppose $\Lambda\in X^*$ and $||\Lambda||=1$. Then $$\Lambda v=\int\Lambda f(t)\,d\mu(t)\le\int||\Lambda||_{X^*}||f(t)||_X\,d\mu(t)=\int||f(t)||\,d\mu(t).$$Since this holds for every such $\Lambda$, Hahn-Banach shows that $||v||\le\int||f||\,d\mu$.

To give a more general picture, that does not use the special form of the $2$-norm being induced by the scalar product, we will show:

Proposition. Suppose $X$ is a Banach space, $(S,\mathcal A, \mu)$ a measure space, and $f \colon S \to X$ is integrable. Then we have $$ \def\norm#1{\left\|#1\right\|}\norm{\int_S f \, d\mu} \le \int_S \norm {f(s)} \, d\mu(s) $$

Proof. We use the definition of the integral. If $f = \sum_i x_i\chi_{A_i}$ is a simple function, where the $A_i$ are disjoint, then $\norm{f(s)} = \sum_i \norm{x_i} \chi_{A_i}(s)$ and hence \begin{align*} \norm{\int_S \sum_{i} x_i \chi_{A_i}d\mu} &= \norm{\sum_i \mu(A_i)x_i}\\ &\le \sum_i \mu(A_i)\norm{x_i}\\ &= \int_S \sum_{i}\norm{x_i}\chi_{A_i} d\mu\\ &= \int_S \norm{f(s)}\, d\mu(s) \end{align*} If $f$ is integrable, choose simple functions $f_n$ such that $f_n \to f$ almost everywhere and $\lim_n\int_S f \, d\mu = \int_S f_n \, d\mu$ we have \begin{align*} \norm{\int_S f\, d\mu} &= \lim_n \norm{\int_S f_n\, d\mu}\\ &\le \lim_n \int_S \norm{f_n(s)}\, d\mu(s)\\ &= \int_S \norm{f(s)}\, d\mu(s) \end{align*}