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

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


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...

Tags:

Inequality

Integration

Real Analysis

Definite Integrals

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