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