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The problem shown is the integral:

\[
\int_0^5 e^{\lfloor x \rfloor} \, dx
\]

where \( \lfloor x \rfloor \) denotes the greatest integer less than or equal to \( x \). Let's break it down.

### Step-by-step breakdown:
Since \( \lfloor x \rfloor \) takes integer values over different intervals of \( x \), we can split the integral according to the integer values of \( x \) over the interval \( [0, 5] \). The function \( e^{\lfloor x \rfloor} \) changes at each integer point.

1. For \( 0 \leq x < 1 \), \( \lfloor x \rfloor = 0 \), so \( e^{\lfloor x \rfloor} = e^0 = 1 \).
2. For \( 1 \leq x < 2 \), \( \lfloor x \rfloor = 1 \), so \( e^{\lfloor x \rfloor} = e^1 = e \).
3. For \( 2 \leq x < 3 \), \( \lfloor x \rfloor = 2 \), so \( e^{\lfloor x \rfloor} = e^2 \).
4. For \( 3 \leq x < 4 \), \( \lfloor x \rfloor = 3 \), so \( e^{\lfloor x \rfloor} = e^3 \).
5. For \( 4 \leq x < 5 \), \( \lfloor x \rfloor = 4 \), so \( e^{\lfloor x \rfloor} = e^4 \).

Now, we compute the integral over each interval:

\[
\int_0^1 1 \, dx = 1
\]
\[
\int_1^2 e \, dx = e \cdot (2 - 1) = e
\]
\[
\int_2^3 e^2 \, dx = e^2 \cdot (3 - 2) = e^2
\]
\[
\int_3^4 e^3 \, dx = e^3 \cdot (4 - 3) = e^3
\]
\[
\int_4^5 e^4 \, dx = e^4 \cdot (5 - 4) = e^4
\]

Adding these up:

\[
1 + e + e^2 + e^3 + e^4
\]

Thus, the value of the integral is:

\[
1 + e + e^2 + e^3 + e^4
\]

### Answer:
This corresponds to the option:

\[
\boxed{e^5 - 1} \quad \text{(Option C)}
\]

Would you like more details or have any questions?

### Related Questions:
1. What is the difference between \( \lfloor x \rfloor \) and \( \lceil x \rceil \)?
2. How do step functions like \( e^{\lfloor x \rfloor} \) affect the behavior of integrals?
3. Can this type of problem be generalized to other functions of \( \lfloor x \rfloor \)?
4. What is the significance of the exponential function \( e^x \) in calculus?
5. How do we approach integrals where the function is piecewise or discontinuous?

#### Tip:
For integrals involving piecewise or discontinuous functions, always split the integral into intervals where the function behaves regularly.

∫ (from 0 to 5) e^[⌊x⌋] dx, where ⌊x⌋ denotes the greatest integer less than or equal to x.

Learn how to evaluate the definite integral ∫ (from 0 to 5) e^[⌊x⌋] dx, where ⌊x⌋ denotes the greatest integer less than or equal to x. This problem involves the properties of integrals and piecewise integration of discontinuous functions. Suitable for college calculus students.

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