Thinking about this $$\left|\int_a^bf(x)\,\mathrm dx\right|\le\int_a^b|f(x)|\,\mathrm dx.$$
Why is this proof wrong? Be $f$ a continuous and integrable function, with primitive $F$, hence
$$\left|\int_a^bf(x)\,\mathrm dx\right| = \left|F(b) - F(a) \right| \leq \left|F(b) \right| - \left| F(a) \right| = \int_a^b \left| f(x) \right|\,\mathrm dx$$
Where I used the fact that $|a - b| = |a + (-b)| \leq |a| + |(-b)|$.
for $a\leq b$ let the $x_0=a ,x_n=b $ choose $n-1$ points in $[a,b]$ and call them $x_2 ,x_3, \dots x_{n-1}$ such that $x_k > x_{k-1}$ then make the interval $I_k:= [x_{k-1}, x_k]$ for $0 \leq k \leq n$ define the norm of that partition by $|P|= \sup\limits_{k=1}^n \{ x_{k}- x_{k-1} \}$, from every interval $I_k$ choose any point and label it $t_k$ then we can define the integral $\int_a ^b f(t) dt $ if $\forall \epsilon >0 \ \exists \delta>0$ such that if $|P|<\delta$ then $\left| \sum\limits_{k=1}^n f(t_k) (x_{k}- x_{k-1}) - L \right| < \epsilon $ for some $L \in \mathbb{R}$ that number $L $ is $\int_a ^b f(t) dt$ if $a> b $ we define $\int_a ^b f(t) dt=-\int_b ^a f(t) dt$ with that being said
$$\left| \sum\limits_{k=1}^n f(t_k)(x_{k}- x_{k-1}) \right| \leq \sum\limits_{k=1}^n |f(t_k)| (x_{k}- x_{k-1}) \ \ \ \ \ \ \ \text{by triangle inequality }$$
then $\left|\int_a ^b f(t) \right| \leq\int_a ^b |f(t)| dt$