Integral Inequality Absolute Value: $\left| \int_{a}^{b} f(x) g(x) \ dx \right| \leq \int_{a}^{b} |f(x)|\cdot |g(x)| \ dx$
Solution 1:
The big idea here is this:
First: it is enough to show that $$ \left\lvert\int_a^b f(x)\,dx\right\rvert\leq\int_a^b\lvert f(x)\rvert dx, $$ since you can replace $f(x)$ by $f(x)\cdot g(x)$ to get the desired result.
Now, notice that $$ -\lvert f(x)\rvert\leq f(x)\leq \lvert f(x)\rvert $$ for all $x$; hence $$ -\int_a^b\lvert f(x)\rvert\,dx\leq \int_a^b f(x)\,dx\leq\int_a^b\lvert f(x)\rvert\,dx. $$ Can you finish it from here?