Finding the Integral of $||x||$ from 0 to 100

[anemone]

Let $||x||$ be the distance of $x$ from the nearest integer, e.g. $||2.6||=0.4$, or $||1.2||=0.2$.

Determine $\displaystyle \int_{0}^{100} ||x||\,dx$.

 

[kaliprasad]

Spoiler

The integral from x to x +t where t =.5 is .25 as we need to integrate t from 0 to .5 x is integer
The integral from x-t to x where t =.5 is .25

So integral from x to x+ 1 for integer x = .5
So integral from 0 to 100 is 50

 

[MarkFL]

My solution:

Spoiler

\(\displaystyle \int_{0}^{100}||x||\,dx=100\left(\int_0^{\frac{1}{2}}x\,dx+\int_{\frac{1}{2}}^{1}1-x\,dx\right)=\)

\(\displaystyle 100\left(\left[\frac{x^2}{2}\right]_0^{\frac{1}{2}}+\left[x-\frac{x^2}{2}\right]_{\frac{1}{2}}^{1}\right)=\)

\(\displaystyle 100\left(\frac{1}{8}+1-\frac{1}{2}-\frac{1}{2}+\frac{1}{8}\right)=100\cdot\frac{1}{4}=25\)

edit: a simpler computation:

Spoiler

\(\displaystyle \int_{0}^{100}||x||\,dx=100\left(\int_0^{\frac{1}{2}}x\,dx+\int_{\frac{1}{2}}^{1}1-x\,dx\right)=\)

\(\displaystyle 100\left(\int_0^{\frac{1}{2}}x\,dx+\int_{0}^{\frac{1}{2}}\frac{1}{2}-x\,dx\right)=\)

\(\displaystyle 50\int_0^{\frac{1}{2}}\,dx=25\)

 

[anemone]

Spoiler

The integral from x to x +t where t =.5 is .25 as we need to integrate t from 0 to .5 x is integer
The integral from x-t to x where t =.5 is .25

So integral from x to x+ 1 for integer x = .5
So integral from 0 to 100 is 50

Hey kali, I think both the integral should give a 0.125 but not 0.25 so your answer is off since it is twice as large as the actual answer.:(

My solution:

Spoiler

\(\displaystyle \int_{0}^{100}||x||\,dx=100\left(\int_0^{\frac{1}{2}}x\,dx+\int_{\frac{1}{2}}^{1}1-x\,dx\right)=\)

\(\displaystyle 100\left(\left[\frac{x^2}{2}\right]_0^{\frac{1}{2}}+\left[x-\frac{x^2}{2}\right]_{\frac{1}{2}}^{1}\right)=\)

\(\displaystyle 100\left(\frac{1}{8}+1-\frac{1}{2}-\frac{1}{2}+\frac{1}{8}\right)=100\cdot\frac{1}{4}=25\)

edit: a simpler computation:

Spoiler

\(\displaystyle \int_{0}^{100}||x||\,dx=100\left(\int_0^{\frac{1}{2}}x\,dx+\int_{\frac{1}{2}}^{1}1-x\,dx\right)=\)

\(\displaystyle 100\left(\int_0^{\frac{1}{2}}x\,dx+\int_{0}^{\frac{1}{2}}\frac{1}{2}-x\,dx\right)=\)

\(\displaystyle 50\int_0^{\frac{1}{2}}\,dx=25\)

Well done, MarkFL! I especially like your second simpler computation and thanks for participating!

 

[Deveno]

Some simple observations:

[sp]$\|x\|$ is periodic, with a period of 1. So it suffices to integrate from 0 to 1.

Also, the graph of $f(x) = \|x\|$ on the interval $[0,1]$ is symmetric about the line $x = \frac{1}{2}$, and $f$ is non-negative on $[0,1]$, so it suffices to integrate from 0 to $\frac{1}{2}$. But on this interval, said integral is just the area of a triangle with base $\frac{1}{2}$ and height $\frac{1}{2}$ (with area $\frac{1}{8}$). It follows, then, that the integral is $\frac{200}{8} = 25$.[/sp]

 

[anemone]

Some simple observations:

[sp]$\|x\|$ is periodic, with a period of 1. So it suffices to integrate from 0 to 1.

Also, the graph of $f(x) = \|x\|$ on the interval $[0,1]$ is symmetric about the line $x = \frac{1}{2}$, and $f$ is non-negative on $[0,1]$, so it suffices to integrate from 0 to $\frac{1}{2}$. But on this interval, said integral is just the area of a triangle with base $\frac{1}{2}$ and height $\frac{1}{2}$ (with area $\frac{1}{8}$). It follows, then, that the integral is $\frac{200}{8} = 25$.[/sp]

Hi Deveno,

What a great observation! Honestly speaking, I didn't think of relating it to a triangle and then solved it geometrically, thanks for participating and sharing your solution with us. I appreciate that!

 

[kaliprasad]

Hey kali, I think both the integral should give a 0.125 but not 0.25 so your answer is off since it is twice as large as the actual answer.:(

Well done, MarkFL! I especially like your second simpler computation and thanks for participating!

Thanks. The base is 1/2 height is 1/2 so area is 1/2 * 1/2 * 1/2 = 1/8 and I did a mistake in calculation. I realized it after seeing Marks's Ans.