> [!NOTE] Theorem > Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. Let $f:[a,b]\to \mathbb{R}$ be [[Darboux Integrable Function|Darboux integrable]]. Then $|f|$ is integrable and $\left\lvert \int_{a}^{b} f(x) \, dx \right\rvert \leq \int_{a}^{b} |f(x)| \, dx \tag{\dagger}$where $x \mapsto |x|$ denotes the [[Absolute Value Function|absolute value function]]. **Proof**: By [[Absolute Value Function is Continuous Everywhere]], $\phi(x)=|x|$ is continuous on $\mathbb{R}.$ By [[Continuous Real Function of Darboux Integrable Function is Darboux Integrable]], $|f|$ is indeed integrable. Now $f\leq |f|$ and $-f\leq |f|$ so by [[Monotonicity of Darboux Integral]], $ \int_a^b f(x) d x \leq \int_a^b|f(x)| d x \quad \text { and } \quad-\int_a^b f(x) d x \leq \int_a^b|f(x)| d x $which gives $\max \left\{ \int_{a}^{b} f(x) d x, \;-\int_{a}^{b} f(x) d x \right\} \leq \int_a^b|f(x)| d x$so $({\dagger})$ is true by [[Absolute Value Function in Terms of Max Function]].