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
$$\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right|
$$
 A: 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...
A: If $f$ is a real Riemann-integrable function, this inequality is true. (and if $f$ is complex Riemann-integrable, then this inequality holds.)
By properties of the modulus function, we have
$$-|f(x)|\le f(x)\le |f(x)|.$$
Since $f$ is continous, $|f|$ is also continous, and hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-\int_a^b |f(x)|dx\le \int_a^bf(x)dx\le \int_a^b|f(x)|dx.$$
From this, we have
$$\left| \int_a^b f(x)dx\right|\le \int_a^b|f(x)|dx.$$
A: 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. 
A: 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.
