Question

If $[\cdot]$ denotes the greatest integer function, then $\int_{0}^{1000} e^{x - \lfloor x \rfloor} dx =$

• A. $\frac{e^{1000} - 1}{1000}$
• B. $1000(e - 1)$
• C. $\frac{e^{1000} - 1}{e - 1}$
• D. $\frac{e - 1}{1000}$

Hint:

When dealing with integrals involving the greatest integer function, look for periodicity in the integrand. The property $\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for periodic functions $f(x)$ with period $T$ can simplify the evaluation significantly. Also, remember that for $0 \le x<1$, $\lfloor x \rfloor = 0$.

Answer 1

The Correct Option is B

Step 1: Identify the periodicity of the integrand.
Let $f(x) = e^{x - \lfloor x \rfloor}$. The term $x - \lfloor x \rfloor = \{x\}$ represents the fractional part of $x$, which is periodic with a period of 1. Therefore, $f(x)$ is also periodic with a period $T = 1$.
Step 2: Use the property of integrals of periodic functions.
For a periodic function $f(x)$ with period $T$, we have $\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$, where $n$ is an integer. In our case, $n = 1000$ and $T = 1$. $$\int_{0}^{1000} e^{x - \lfloor x \rfloor} dx = 1000 \int_{0}^{1} e^{x - \lfloor x \rfloor} dx$$
Step 3: Evaluate the integral over one period.
For $0 \le x<1$, the greatest integer function $\lfloor x \rfloor = 0$. Thus, $x - \lfloor x \rfloor = x$. $$\int_{0}^{1} e^{x - \lfloor x \rfloor} dx = \int_{0}^{1} e^{x} dx$$ Evaluating this integral: $$\int_{0}^{1} e^{x} dx = [e^x]_{0}^{1} = e^1 - e^0 = e - 1$$
Step 4: Combine the results.
Substituting the value of the integral over one period back into the equation from Step 2: $$\int_{0}^{1000} e^{x - \lfloor x \rfloor} dx = 1000 (e - 1)$$ Thus, the value of the integral is $\boxed{1000(e - 1)}$.