Question

If $\{.\}$ represents the fractional part function, then the value of $\int_{0}^{1000} e^{\{x\}} dx$ is equal to ____.

• A. 1000e - 1
• B. 1000e
• C. 1000 (e - 1)
• D. none of these

Answer 1

Answer: (C)

1000 (e - 1)

Answer 2

To evaluate the integral $\int_{0}^{1000} e^{\{x\}} dx$, we use the properties of the fractional part function and definite integrals.

The fractional part function $\{x\}$ is defined as $\{x\} = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. The function $\{x\}$ is periodic with a period of 1. This means $\{x+1\} = \{x\}$. Consequently, the function $f(x) = e^{\{x\}}$ is also periodic with a period of 1. $e^{\{x+1\}} = e^{\{(x+1) - \lfloor x+1 \rfloor\}} = e^{\{x+1 - (\lfloor x \rfloor + 1)\}} = e^{\{x - \lfloor x \rfloor\}} = e^{\{x\}}$.

For a periodic function $f(x)$ with period $T$, the property of definite integrals states that $\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$. In this problem, $f(x) = e^{\{x\}}$, the period $T=1$, and the upper limit of integration is 1000, which can be written as $1000 \times 1$. So, $n=1000$.

Applying this property:

$$\int_{0}^{1000} e^{\{x\}} dx = 1000 \int_{0}^{1} e^{\{x\}} dx$$

Now, we need to evaluate the integral over one period, i.e., $\int_{0}^{1} e^{\{x\}} dx$. For $x \in [0, 1)$, the fractional part $\{x\}$ is simply equal to $x$. So, the integral becomes:

$$\int_{0}^{1} e^{\{x\}} dx = \int_{0}^{1} e^{x} dx$$

This is a standard integral:

$$\int_{0}^{1} e^{x} dx = [e^x]_{0}^{1} = e^1 - e^0 = e - 1$$

Substitute this result back into the main expression:

$$\int_{0}^{1000} e^{\{x\}} dx = 1000 (e - 1)$$