Show $|\int^{b}_{a} f(x) dx | \leq \int^{b}_{a} |f(x)| dx $

Question

Show:

$$|\int^{b}_{a} f(x) dx | \leq \int^{b}_{a} |f(x)|$$

Hint: $-|f(x)| \leq f(x) \leq |f(x)|$ holds for all $x \in dom(f)$

I tried rewriting the integral as a riemann sum but then I just have one big absolute limit and it doesn't really help.

$$|\int^{b}_{a} f(x) dx | = |\lim_{n \to \infty} \sum^{n}_{i=1}f(x_{i}^{*})\Delta x|$$

Answer 1

You know that $-|f (x)| \le f (x) \le |f (x)|$.

Integration conserves inequalities (if b > a) so $-\int_a^b |f (x)| \le \int_a^b f (x) \le \int_a^b |f (x)|$.

Then, by definition, this means that $|\int_a^b f (x)| \le \int_a^b |f (x)|$

Answer 2

An alternative way to show it (without following the hint) would be:

Write $f$ as the sum of its positive and negative parts $$f(x)=\underbrace{\max\{f(x),0\}}_{=:f^+(x)}+\underbrace{\min\{f(x),0\}}_{=:f^-(x)},$$ and then one has, using the linearity of the integral and the triangular inequality, \begin{align} \left|\int_a^bf(x)\,dx\right|=&\left|\int_a^bf^+(x)\,dx+\int_a^bf^-(x)\,dx\right|\\ \leq&\left|\int_a^bf^+(x)\,dx\right|+\left|\int_a^bf^-(x)\,dx\right|\\ =&\int_a^bf^+(x)\,dx-\int_a^bf^-(x)\,dx\\ =&\int_a^b\underbrace{(f^+(x)-f^-(x))}_{=|f(x)|}\,dx. \end{align}