To solve \(\int_{0}^{1} (1-x)e^{-\frac{1}{2}x} \, dx\), we use integration by parts. Let \(u = 1-x\) and \(dv = e^{-\frac{1}{2}x} \, dx\).

Then \(du = -dx\) and \(v = -2e^{-\frac{1}{2}x}\).

Applying integration by parts:

\(\int u \, dv = uv - \int v \, du\)

\(= (1-x)(-2e^{-\frac{1}{2}x}) \bigg|_{0}^{1} + 2 \int_{0}^{1} e^{-\frac{1}{2}x} \, dx\)

Evaluating the first term:

\(= [-2(1-x)e^{-\frac{1}{2}x}] \bigg|_{0}^{1}\)

\(= [-2(0)e^{-\frac{1}{2}(1)} + 2(1)e^{0}]\)

\(= [0 + 2] = 2\)

Now, evaluate the second integral:

\(2 \int_{0}^{1} e^{-\frac{1}{2}x} \, dx = 2 \left[ -2e^{-\frac{1}{2}x} \right]_{0}^{1}\)

\(= 2 \left[ -2e^{-\frac{1}{2}} + 2 \right]\)

\(= -4e^{-\frac{1}{2}} + 4\)

Combining both parts:

\(2 - 4e^{-\frac{1}{2}} + 4 = 4 - 4e^{-\frac{1}{2}}\)

Thus, the integral evaluates to:

\(4e^{-\frac{1}{2}} - 2\)