To solve the given equation, let's first rewrite it for clarity:

\[ (1 + 2 + 3 + \ldots + 2011)x = 2012 \cdot 2013 \]

We need to find the sum of the series \(1 + 2 + 3 + \ldots + 2011\). This is an arithmetic series where the first term \(a = 1\) and the last term \(l = 2011\), with the number of terms \(n = 2011\).

The sum \(S_n\) of the first \(n\) terms of an arithmetic series is given by:

\[ S_n = \frac{n}{2} (a + l) \]

Substituting the values:

\[ S_{2011} = \frac{2011}{2} (1 + 2011) \]
\[ S_{2011} = \frac{2011}{2} \cdot 2012 \]
\[ S_{2011} = 2011 \cdot 1006 \]

So, the equation becomes:

\[ (2011 \cdot 1006)x = 2012 \cdot 2013 \]

To solve for \(x\), we divide both sides by \(2011 \cdot 1006\):

\[ x = \frac{2012 \cdot 2013}{2011 \cdot 1006} \]

Let's simplify this fraction. Notice that:

\[ 2012 = 2011 + 1 \]
\[ 2013 = 2011 + 2 \]

So the fraction becomes:

\[ x = \frac{(2011 + 1)(2011 + 2)}{2011 \cdot 1006} \]

Expanding the numerator:

\[ x = \frac{2011^2 + 3 \cdot 2011 + 2}{2011 \cdot 1006} \]

Since \(2011^2\) and \(2011\) are common factors, we can simplify further:

\[ x = \frac{2011(2011 + 3) + 2}{2011 \cdot 1006} \]
\[ x = \frac{2011^2 + 3 \cdot 2011 + 2}{2011 \cdot 1006} \]

This simplifies to:

\[ x = \frac{2011^2 + 3 \cdot 2011 + 2}{2011 \cdot 1006} \]

Since the numerator and denominator share the common factor \(2011\), we can simplify:

\[ x = \frac{2011 + 3 + \frac{2}{2011}}{1006} \]

Given that \(\frac{2}{2011}\) is a very small number, it can be approximated to zero for simplicity:

\[ x \approx \frac{2011 + 3}{1006} \]
\[ x \approx \frac{2014}{1006} \]
\[ x \approx 2 \]

Thus, the value of \(x\) is approximately 2.