How to prove $|\int^{b}_{a}f(x)dx| \leq \int^{b}_{a} |f(x)|dx$ using riemann sum. $$|\int^{b}_{a}f(x)dx| \leq \int^{b}_{a} |f(x)|dx$$
We know for a fact that $a \leq |a|$.
Also, prove using riemann sum.
$$\lim_{n \to \infty} \sum^{n}_{i=1}f(x^*)\Delta x = \int^{b}_{a}f(x)dx$$
Thus,
$$|\lim_{n \to \infty} \sum^{n}_{i=1}f(x^*)\Delta x| = |\int^{b}_{a}f(x)dx|$$
By limit laws, we can pull the limit outside.
$$ \lim_{n \to \infty} |\sum^{n}_{i=1}f(x^*)\Delta x| = |\int^{b}_{a}f(x)dx|$$
Thus,
$$ \lim_{n \to \infty} |\sum^{n}_{i=1}f(x^*)\Delta x| = \int^{b}_{a}|f(x)dx|$$
Have I done all the steps right? I want to show that its $\leq$ but I cannot think of a way how?

By the triangle inequality we have,
$$|b_1+b_2| \leq |b_1|+|b_2|$$
This implies
$$|b_1+b_2+b_3| \leq |b_1+b_2|+|b_3|$$
$$\leq |b_1|+|b_2|+|b_3|$$
Etc.
It follows that,
$$\sum_{i=1}^{n} |b_i| \geq |\sum_{i=1}^{n} b_i|$$
If you want to more rigorously show this use induction.
Now choose $b_i=f(x^{*}) \Delta x$. Assuming $b>a$, then $\Delta x>0$. Then we have,
$$\sum_{i=1}^{n} |f(x^{*})| \Delta x \geq |\sum_{i=1}^{n} f(x^{*}) \Delta x|$$
In the limit, as $n \to \infty$ then,
$$\int_{a}^{b} |f(x)| dx \geq |\int_{a}^{b} f(x) dx |$$
