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

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.

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

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    Just integrate the inequality given by the hint – 2017-01-24
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    You are almost there. Just note $|\int^{b}_{a} f(x) dx | = |\lim_{n \to \infty} \sum^{n}_{i=1}f(x_{i}^{*})\Delta x|\le \lim_{n \to \infty} \sum^{n}_{i=1}|f(x_{i}^{*})|\Delta x=\int^{b}_{a} |f(x)|dx$ – 2017-01-24
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    Is it included in the question that $|f|$ is integrable given that $f$ is? Otherwise one should give an argument for that. – 2017-01-24
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    It is integrable and continous – 2017-01-24
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    @User1006 how do you know that the inequality you proposed holds? How can you take absolute value of a limit and say it's less then absolute value of function – 2017-01-24
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    What are you treating at f(x)? The entire limit? – 2017-01-24
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    @K Split X, there is $|\lim_{n\to\infty}a_n|=\lim_{n\to\infty}|a_n|$, which can be proved easily – 2017-01-24
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    Related: https://math.stackexchange.com/questions/1711885 – 2018-11-27

2 Answers 2

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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 $|\int_a^b f (x)| \le \int_a^b |f (x)|$

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    How did you go from the second step to the third step? – 2017-01-24
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    How does $-\int^{b}_{a}|f(x)|$ turn into $|\int^{b}_{a}f(x)|$ – 2017-01-24
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    If $-M \le a \le M $ and $M \ge 0$, then $|a| \le M $. That's maybe not the definition of absolute value, but it is straightforward. If you want to prove it, just prove it's true for every negative $a $, and that it's true for every positive $a $. – 2017-01-24
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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}

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    (+1) This is the approach I would have suggested. Clear and concise. – 2017-01-24