Prove $|\int_{a}^{b}f(x)dx| \le \int_{a}^{b}|f(x)|dx $ I want to prove that $|\int_{a}^{b}f(x)dx| \le \int_{a}^{b}|f(x)|dx $
We know that $-|f(x)| \le f(x) \le |f(x)|$, so by linearity, we get:
$$\int_{a}^{b}-|f(x)|dx \le \int_{a}^{b}f(x)dx \le \int_{a}^{b}|f(x)|dx$$
And:
$$-\int_{a}^{b}|f(x)|dx \le \int_{a}^{b}f(x)dx \le \int_{a}^{b}|f(x)|dx$$
But how can I conclude that the statement is correct from here?
Thanks!
 A: Note that for any $a \in \mathbb{R}$, if we have $-a \le x \le a$ we get $|x| \le a$
So $ |\int_{a}^{b}f(x)dx| \le \int_{a}^{b}|f(x)|dx$
A: we know that $$\left|\int_a^b f(x)dx\right|=\cases{\int_a^b f(x)dx, \text{ if the integral}\ge0 \\ -\int_a^b f(x)dx, \text{ if the integral}<0}$$
but $$\pm \int_a^b f(x)dx\le \int_a^b | f(x)| dx$$
so we get $$\left|\int_a^b f(x)dx\right|\le  \int_a^b | f(x)| dx$$
A: First, take a function:
$$f_1(x)\ge0\forall x\in[a,b]$$
it is then clear that:
$$|f_1(x)|=f_1(x)$$
and so:
$$\int_a^b|f_1(x)|dx=\int_a^bf_1(x)dx$$
and then it follows that:
$$\left|\int_a^bf_1(x)dx\right|=\int_a^b|f_1(x)|dx$$

Now lets make a new function, $a<c<b$:
$$f(x)=\begin{cases}f(x)\ge0&a\le x<c\\f(x)\le0&c\le x\le b\end{cases}$$
now we can say that:
$$\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx$$
if we assign values to these we can use them later:
$$F=F_1+F_2$$
now notice that $F_1\ge0,F_2\le0$:
$$\int_a^b|f(x)|dx=|F_1|+|F_2|$$
$$\left|\int_a^bf(x)dx\right|=|F_1+F_2|$$
now since we know the values of $F_1,F_2$ it is a given that:
$$|F_1+F_2|\le|F_1|+|F_2|$$
