$|int_a^b f(x)dx| leq |int_a^b |f(x)| dx| $?

Question

For $a>b$ Does the following hold:
$|int_a^b f(x)dx| leq |int_a^b |f(x)| dx| $ ?
At $"leq"$ I have used the triangular inequality for integrals:
If $f:Ito mathbb{R} $ is Riemann integrable over $Isubseteq mathbb{R}$ and $a,b in I$ with $a < b$, then $|int_a^b f(x)dx| leq int_a^b |f(x)| dx $.

Kavi Rama Murthy · Answer

When $a>b$ we define $int_a^{b}f(x)dx$ as $- int_b^{a}f(x)dx$. So the inequality is true and it reduces to the case $a<b$.
To be explicit we have $|int_a^{b} f(x)dx| =|int_b^{a}f(x)dx| leq int_b^{a}|f(x)|dx=|int_a^{b}|f(x)|dx|$

$|int_a^b f(x)dx| leq |int_a^b |f(x)| dx| $?

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