$\int_0^1 x\left|x-\frac{1}{2}\right| \mathrm{d} x=$

The integrand $x\left|x-\frac{1}{2}\right|$ has a discontinuity in its derivative at $x = \frac{1}{2}$, requiring integration over two intervals.

For $x < \frac{1}{2}$, $\left|x-\frac{1}{2}\right| = \frac{1}{2} - x$. For $x \geq \frac{1}{2}$, $\left|x-\frac{1}{2}\right| = x - \frac{1}{2}$. The integral splits accordingly:

$\int_0^1 x\left|x-\frac{1}{2}\right| dx = \int_0^{1/2} x\left(\frac{1}{2} - x\right) dx + \int_{1/2}^1 x\left(x - \frac{1}{2}\right) dx$

The first integral evaluates as:

$\int_0^{1/2} \left(\frac{x}{2} - x^2\right) dx = \left[\frac{x^2}{4} - \frac{x^3}{3}\right]_0^{1/2} = \frac{1}{16} - \frac{1}{24} = \frac{1}{48}$

The second integral yields:

$\int_{1/2}^1 \left(x^2 - \frac{x}{2}\right) dx = \left[\frac{x^3}{3} - \frac{x^2}{4}\right]_{1/2}^1 = \left(\frac{1}{3} - \frac{1}{4}\right) - \left(\frac{1}{24} - \frac{1}{16}\right) = \frac{1}{12} + \frac{1}{48} = \frac{5}{48}$

Summing both parts gives $\frac{1}{48} + \frac{5}{48} = \frac{6}{48} = \frac{1}{8}$.

Final result: $\boxed{\frac{1}{8}}$