Question:

Step 1: Break down the integral using the periodicity of the greatest integer and fractional part functions. Since the greatest integer function $[x]$ is constant on each interval $[n, n+1)$ where $n$ is a non-negative integer, we can split the integral from $0$ to $10$ into ten subintervals: $$\int_{0}^{10} [x]^3 \{x\} \, dx = \sum_{n=0}^{9} \int_{n}^{n+1} [x]^3 \{x\} \, dx$$ Step 2: Identify the values of $[x]$ and $\{x\}$ on each subinterval. For $x \in [n, n+1)$ where $n$ is a non-negative integer: - The greatest integer function: $[x] = n$ - The fractional part function: $\{x\} = x - n$ Step 3: Evaluate the integral on a single subinterval $[n, n+1)$. Substituting the expressions from Step 2: $$\int_{n}^{n+1} n^3(x - n) \, dx$$ Let $t = x - n$, so $dt = dx$. When $x = n$, $t = 0$; when $x = n+1$, $t = 1$: $$n^3 \int_{0}^{1} t \, dt = n^3 \left[\frac{t^2}{2}\right]_{0}^{1} = n^3 \cdot \frac{1}{2} = \frac{n^3}{2}$$ Step 4: Sum the contributions from all subintervals. $$\int_{0}^{10} [x]^3 \{x\} \, dx = \sum_{n=0}^{9} \frac{n^3}{2} = \frac{1}{2} \sum_{n=0}^{9} n^3$$ Step 5: Apply the formula for the sum of cubes. The formula for the sum of cubes is: $$\sum_{n=1}^{N} n^3 = \left[\frac{N(N+1)}{2}\right]^2$$ For $n = 0$ to $9$, note that the $n=0$ term contributes $0$, so: $$\sum_{n=0}^{9} n^3 = \sum_{n=1}^{9} n^3 = \left[\frac{9 \cdot 10}{2}\right]^2 = (45)^2 = 2025$$ Step 6: Calculate the final answer. $$\int_{0}^{10} [x]^3 \{x\} \, dx = \frac{1}{2} \cdot 2025 = \frac{2025}{2}$$ **Final Answer:** The value of $\int_{0}^{10} [x]^3 \{x\} \, dx = \boxed{\frac{2025}{2}}$ This corresponds to **Option 2**.