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The problem asks to compute the line integral:

\[
\int_L (x + y + z) \, dl
\]

where \( L \) is the line segment connecting points \( A(0,0,0) \) and \( B(1,1,1) \).

### Solution:

The line segment \( L \) between \( A(0,0,0) \) and \( B(1,1,1) \) can be parametrized as:

\[
\mathbf{r}(t) = (t, t, t), \quad t \in [0, 1]
\]

This is because \( x = t \), \( y = t \), and \( z = t \) for \( t \in [0, 1] \).

Now, we can calculate the differential line element \( dl \). The magnitude of the derivative of \( \mathbf{r}(t) \) is:

\[
dl = \|\mathbf{r}'(t)\| \, dt = \sqrt{1^2 + 1^2 + 1^2} \, dt = \sqrt{3} \, dt
\]

Now, substitute the values of \( x = t \), \( y = t \), and \( z = t \) into the integrand:

\[
x + y + z = t + t + t = 3t
\]

Thus, the integral becomes:

\[
\int_0^1 (3t) \cdot \sqrt{3} \, dt = 3\sqrt{3} \int_0^1 t \, dt
\]

The integral of \( t \) is:

\[
\int_0^1 t \, dt = \frac{t^2}{2} \Big|_0^1 = \frac{1}{2}
\]

So, the result is:

\[
3\sqrt{3} \cdot \frac{1}{2} = \frac{3\sqrt{3}}{2}
\]

### Final Answer:

\[
\frac{3\sqrt{3}}{2}
\]

Let me know if you need further details or have any questions!

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Here are 5 related questions to expand on this topic:
1. How would the integral change if the line segment was not linear but a curve?
2. Can you parametrize a different line integral between two arbitrary points?
3. What is the physical interpretation of a line integral in this context?
4. How do you compute a surface integral over a plane or curved surface?
5. What is the difference between a line integral and a regular definite integral?

**Tip:** For line integrals, always ensure you correctly parametrize the path and compute the differential length \( dl \) accordingly.

Обчислити ∫(x + y + z)dl, де L - відрізок прямої між A(0,0,0) та B(1,1,1).

Learn how to compute the line integral ∫(x + y + z)dl over the line segment between points (0,0,0) and (1,1,1). This problem involves parametrizing the line and integrating with respect to the path length. Suitable for university-level students studying vector calculus.

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