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.