For a real number x let `[x]` denote the largest integer less than or equal to x and `{x}=x-[x]`. The possible integer value of n for which `int_(1)^(n)[x]{x}dx` exceeds 2013 is

Options

A. 63
B. 64
C. 90
D. 91

Explanation

To solve the problem, we need to evaluate the integral

\[ I(n) = \int_{1}^{n} [x] \{x\} \, dx \]

where \([x]\) is the greatest integer less than or equal to \(x\) and \(\{x\} = x - [x]\).

### Step-by-Step Solution:

1. **Understanding the Integral**:
The function \([x]\) is piecewise constant, and \(\{x\}\) is a linear function. For \(x\) in the interval \([k, k+1)\) where \(k\) is an integer, we have:
\[ [x] = k \quad \text{and} \quad \{x\} = x - k \]
Therefore, the integral can be split into intervals:
\[ I(n) = \sum_{k=1}^{n-1} \int_{k}^{k+1} k (x - k) \, dx \]

2. **Calculating the Integral for Each Interval**:
For each \(k\), we compute:
\[ \int_{k}^{k+1} k (x - k) \, dx = k \int_{k}^{k+1} (x - k) \, dx \]
Let \(u = x - k\), then when \(x = k\), \(u = 0\) and when \(x = k+1\), \(u = 1\). Thus:
\[ \int_{k}^{k+1} (x - k) \, dx = \int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{2} \]
Therefore:
\[ \int_{k}^{k+1} k (x - k) \, dx = k \cdot \frac{1}{2} = \frac{k}{2} \]

3. **Summing Over All Intervals**:
Now we sum this from \(k=1\) to \(n-1\):
\[ I(n) = \sum_{k=1}^{n-1} \frac{k}{2} = \frac{1}{2} \sum_{k=1}^{n-1} k = \frac{1}{2} \cdot \frac{(n-1)n}{2} = \frac{(n-1)n}{4} \]

4. **Setting Up the Inequality**:
We need to find \(n\) such that:
\[ I(n) > 2013 \implies \frac{(n-1)n}{4} > 2013 \implies (n-1)n > 8052 \]

5. **Finding Integer Solutions**:
To solve \((n-1)n > 8052\), we can approximate \(n^2 - n - 8052 > 0\). Using the quadratic formula:
\[ n = \frac{1 \pm \sqrt{1 + 4 \cdot 8052}}{2} = \frac{1 \pm \sqrt{32209}}{2} \]
Calculating \(\sqrt{32209} \approx 179.4\), we find:
\[ n \approx \frac{1 + 179.4}{2} \approx 90.2 \]
Thus, we check integer values \(n = 90\) and \(n = 91\).

6. **Checking Values**:
- For \(n = 90\):
\[ I(90) = \frac{89 \cdot 90}{4} = \frac{8010}{4} = 2002.5 \quad (\text{not exceeding } 2013) \]
- For \(n = 91\):
\[ I(91) = \frac{90 \cdot 91}{4} = \frac{8190}{4} = 2047.5 \quad (\text{exceeds } 2013) \]

### Conclusion:
The possible integer value of \(n\) for which the integral exceeds 2013 is:

\[ \boxed{91} \]