# Prove or disprove that $\int_a^b |f(x)|\ \mathrm{d}x\geq \big | \int_a^b f(x)\ \mathrm{d}x\big|$ [closed]

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 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 $$\left| \int_a^b f(x)dx\right|\le \int_a^b|f(x)|dx.$$

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...

• This is exactly the argument I was thinking about, the proper way to do it by just seeing it as a continuous version of the triangle inequality.
– user459879
Dec 8 '18 at 13:44

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.

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 $\vert f(x) \vert$.