Consider negative lengths as positive too.

If we consider negative lengths as positive, we are interested in the expected value of the absolute value of the projection of the segment onto the X-axis. This changes the problem to finding the expected value of \(|\cos(\theta)|\) for a random angle \(\theta\).

### Problem Breakdown

1. **Segment Length**: The segment is 1 meter long.
2. **Projection on the X-Axis**: The projection of the segment onto the X-axis is given by \(\cos(\theta)\).
3. **Absolute Projection**: We now need to find the expected value of the absolute value of the projection, \(|\cos(\theta)|\).
4. **Angle Distribution**: The angle \(\theta\) is uniformly distributed between 0 and \(2\pi\).

### Calculation

The expected value of \(|\cos(\theta)|\) is given by:

\[
E[|\cos(\theta)|] = \frac{1}{2\pi} \int_0^{2\pi} |\cos(\theta)| \, d\theta
\]

Since \(|\cos(\theta)|\) is symmetric around \(\theta = \pi\), we can calculate the integral over the interval \([0, \pi]\) and then double it:

\[
E[|\cos(\theta)|] = \frac{2}{2\pi} \int_0^{\pi} \cos(\theta) \, d\theta
\]

\[
E[|\cos(\theta)|] = \frac{1}{\pi} \int_0^{\pi} \cos(\theta) \, d\theta
\]

### Integral Solution

Let's solve the integral:

\[
\int_0^{\pi} \cos(\theta) \, d\theta = \left[\sin(\theta)\right]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0
\]

The calculation of \(|\cos(\theta)|\) is slightly different, as we consider only the positive values over the range \([0, \pi]\):

\[
\int_0^{\pi} |\cos(\theta)| \, d\theta = \int_0^{\frac{\pi}{2}} \cos(\theta) \, d\theta + \int_{\frac{\pi}{2}}^{\pi} (-\cos(\theta)) \, d\theta
\]

Calculating each part separately:

\[
\int_0^{\frac{\pi}{2}} \cos(\theta) \, d\theta = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1
\]

\[
\int_{\frac{\pi}{2}}^{\pi} -\cos(\theta) \, d\theta = -(\sin(\pi) - \sin\left(\frac{\pi}{2}\right)) = -(0 - 1) = 1
\]

Adding these gives:

\[
\int_0^{\pi} |\cos(\theta)| \, d\theta = 1 + 1 = 2
\]

Thus,

\[
E[|\cos(\theta)|] = \frac{2}{\pi} \times 1 = \frac{2}{\pi}
\]

### Conclusion

The expected length of the projection of a 1-meter segment onto the X-axis, considering all projections as positive, is \(\frac{2}{\pi} \approx 0.6366\) meters.

---

Would you like more details or have any questions on this? Here are some related questions:

1. How does the result change if the segment length is not 1 meter?
2. What is the expected projection length if the segment is restricted to angles between 0 and \(\pi/2\)?
3. How does the expected projection change for a 3D segment?
4. Can you apply this method to find the expected projection on the Y-axis?
5. How does the concept of symmetry simplify this calculation?
6. What is the variance of the projection length?
7. How can this problem be extended to other trigonometric functions?
8. What real-world problems can this mathematical concept be applied to?

**Tip**: When dealing with absolute values in integrals, consider breaking the interval into regions where the function maintains a consistent sign.

Explore the calculation of the expected value of the absolute value of the projection of a 1-meter segment onto the X-axis, considering negative lengths as positive too. Learn the step-by-step solution involving trigonometric functions and integration. Discover how symmetry simplifies this mathematical concept. Suitable for advanced mathematics enthusiasts.

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