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I don't know if there's a name for this but it's sort of like the triangle inequality. Namely:

$$\left| \int f(x) dx \right| \leq \int \left|f(x)\right| dx$$

  1. What is this rule called, if it is even called anything?

  2. Is it even right?

  3. If it is right, how do you prove it?

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  • $\begingroup$ Consider decomposing as $f(x) = f^+(x)+f^-(x)$ where $f^+(x)=\max(0,f(x))$ and $f^-(x)=\min(0,f(x))$ $\endgroup$ – JMoravitz Mar 13 '18 at 16:50
  • $\begingroup$ Do you know that if $f\le g$ over an interval $I$ then $\int_I f\le \int_Ig$? $\endgroup$ – user296113 Mar 13 '18 at 16:50
  • $\begingroup$ proofwiki.org/wiki/Absolute_Value_of_Definite_Integral $\endgroup$ – Kevin Mar 13 '18 at 16:52
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    $\begingroup$ Is it not true for indefinite integrals? $\endgroup$ – user539262 Mar 13 '18 at 16:53
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    $\begingroup$ For the indefinite case, yes. Example: let $f(x) = x^2$. The left-hand side is $F_1(x)=|x^3/3+C_1|$ and the right-hand one is $F_2(x)=x^3/3+C_2$, where $C_1$ and $C_2$ are integration constants. Choose $C_2<C_1$ (and both positive) and you get false for all positive $x$. $\endgroup$ – Cuspy Code Mar 13 '18 at 18:00
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With $f(x)\leq|f(x)|$, then $\displaystyle\int f(x)dx\leq\int|f(x)|dx$. With $f(x)\geq-|f(x)|$, then $\displaystyle\int f(x)dx\geq\int-|f(x)|dx=-\int|f(x)|dx$, so $\left|\displaystyle\int f(x)dx\right|\leq\displaystyle\int|f(x)|dx$.

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  • $\begingroup$ I don't see how you arrived at the last step $\endgroup$ – user539262 Mar 13 '18 at 16:52
  • $\begingroup$ If $a\leq u$ and $a\geq -u$, where $u\geq 0$, then $|a|\leq u$. $\endgroup$ – user284331 Mar 13 '18 at 16:53
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Expanding on my comment, decompose $f$ into its positive and negative parts as $f(x)=f^+(x)+f^-(x)$ where $f^+(x)=\max(0,f(x))$ and $f^-(x)=\min(0,f(x))$.

First, notice that $|f(x)|=f^+(x)-f^-(x)$.

We have then $|\int^a_b f(x)dx|= |\int^a_b f^+(x)dx + \int^a_b f^-(x) dx|\leq |\int^a_b f^+(x)dx|+|\int^a_b f^-(x)dx|$

The inequality above is just the normal triangle inequality. Then, recognizing that the integral on the left is positive or zero and on the right is negative or zero, this continues as

$=\int^a_b f^+(x)dx -\int^a_b f^-(x)dx = \int^a_b f^+(x)-f^-(x)dx = \int^a_b|f(x)|dx$


For indefinite integrals, one has to consider the integration constant that occurs, the $+C$ of $\int f(x)dx = F(x)+C$. Technically, the indefinite integral of a function is a whole family of curves, some of which will be greater than or less than others within its family.

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  • $\begingroup$ Does this imply that for indefinite integrals it is possible to make the inequality false? (is there an example of this?) $\endgroup$ – user539262 Mar 13 '18 at 17:35
  • $\begingroup$ @user539262 it implies that the question of whether one indefinite integral is "less than" or "greater than" another is flawed and doesn't make sense in the first place. What does it mean for an indefinite integral to be greater than zero for instance? Remember that the indefinite integral of a function is by no means unique. We may find an antiderivative of a function but we cannot find the antiderivative of a function since there are many. $\endgroup$ – JMoravitz Mar 13 '18 at 17:40
  • $\begingroup$ I see how you arrive at the triangle inequality for the plus and minus pieces but you lost me after that ("Then, recognizing that the integral ...") -- I don't see how you are incorporating the integral of $|f(x)|$ into this since we're dealing with the sum of two absolute values of its decomposition. $\endgroup$ – user539262 Mar 13 '18 at 17:42
  • $\begingroup$ @user539262 That follows from my second line "First notice that $|f(x)|=f^+(x)-f^-(x)$" as well as the definition of absolute value. The absolute value of a positive number is itself and the absolute value of a negative number is $-1$ times itself. $\endgroup$ – JMoravitz Mar 13 '18 at 17:44
  • $\begingroup$ I understand the $|f(x)|=f^+(x)-f^-(x)$ piece -- I am lost how this is being incorporated into $|\int^a_b f(x)dx| \leq |\int^a_b f^+(x)dx|+|\int^a_b f^-(x)dx|$. Are you saying that $|\int^a_b f^+(x)dx|+|\int^a_b f^-(x)dx| =\int^a_b f^+(x)dx -\int^a_b f^-(x)dx = \int^a_b f^+(x)-f^-(x)dx = \int^a_b|f(x)|dx$? $\endgroup$ – user539262 Mar 13 '18 at 17:47
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It will be easier if you keep in mind that an integral gives a surface. The triangle inequality tells that | x + y | $\leq$ | x | + | y |. Let's now take integral of f(x) between a and b (a $\leq$ b): $\int_a^b$f(x)dx. This gives the sum of the surfaces of the rectangles between a and b that are infinitesimal. Take these surfaces to be x, y, z ... and by using | x + y + z + ...| $\leq$ |x| + |y| + |z| + ... of more generally |$\displaystyle\sum_{i=1}^n x_i$| $\leq$ $\displaystyle\sum_{i=1}^n |x_i|$, you will get a starting idea to prove this inequality.

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