Prove or disprove that $\int_a^b |f(x)|\ \mathrm{d}x\geq \big | \int_a^b f(x)\ \mathrm{d}x\big| $ Let $f$ be a continuous and integrable function over $[a,b]$. Prove or disprove that
$$\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right|
$$
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 
$$\left| \int_a^b f(x)dx\right|\le \int_a^b|f(x)|dx.$$
Think about what your inequality is saying:


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


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