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+Let $f$ be a continuous and integrable function over $[a;b]$, Prove or disprove that :

$\displaystyle\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right| $

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What have you tried? –  Trevor Wilson Jan 22 '13 at 16:47
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4 Answers 4

up vote 3 down vote accepted

If $f$ is real Riemann-integrable function, this inequality is true. (and if $f$ is complex Riemann-integrable, this inequality is true.)

By properties of absolute value, we get $$-|f(x)|\le f(x)\le |f(x)|.$$ Since $f$ is continous, so $|f|$ is also continous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get $$-\int_a^b |f(x)|dx\le \int_a^bf(x)dx\le \int_a^b|f(x)|dx$$ and we get $$\left| \int_a^b f(x)dx\right|\le \int_a^b|f(x)|dx.$$

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The inequality is true. Hints:

For any Riemann sum we get from the usual triangle inequality for the absolute value:

$$\left|\sum_{k=1}^nf(c_i)(x_i-x_{i-1})\right|\leq\sum_{k=1}^n|f(c_i)|(x_i-x_{i-1})\,\,,\,$$

$$\{a=x_0<x_1<...<x_n=b\}\,\,,\,\,c_i\in[x_{i-1},x_1]$$

Pass now to the limit $\,n\to\infty\,$ while the maximal length of the subintervals goes to zero (this is what is done to get the Riemann integral from Riemann sums) and that's all...

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Separate $f$ into it's positive and negative parts, where $f^+$ is always non-negative and $f^-$ is always non-positive, and each are 0 where $f$ is negative and positive respectively. $f$ is obviously $f^+ + f^-$ so $$\int_a^b |f|dx = \int_a^bf^++(-f^-)dx \geq max \{\int_a^bf^+dx , \int_a^b-f^-dx \} \geq |\int_a^bf^++f^-dx |$$

which proves the theorem positively.

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Think about what your inequality is saying:

  1. Imagine coordinates $y$ and $x$.
  2. Imagine some arbitrary curve on that plane $f(x)$.
  3. Imagine integrating $f(x)$ by taking the area under the $x$-axis away from the area above the $x$-axis.
  4. Imagine repeating the integration for $\vert f(x) \vert$.

And you can see that your inequality is true for integrable functions.

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