Is there a way to prove this inequality for integrals?

Question

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?

Answer 1

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

Answer 2

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

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

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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.

Answer 3

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.