To solve \(\int_{0}^{1} (2-x)e^{-2x} \, dx\), we use integration by parts. Let \(u = 2-x\) and \(dv = e^{-2x} \, dx\).
Then \(du = -dx\) and \(v = -\frac{1}{2}e^{-2x}\).
Applying integration by parts:
\(\int u \, dv = uv - \int v \, du\)
\(= (2-x)\left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right)(-1) \, dx\)
\(= -\frac{1}{2}(2-x)e^{-2x} - \frac{1}{2} \int e^{-2x} \, dx\)
Integrate \(\int e^{-2x} \, dx\):
\(\int e^{-2x} \, dx = -\frac{1}{2}e^{-2x} + C\)
Substitute back:
\(-\frac{1}{2}(2-x)e^{-2x} - \frac{1}{4}e^{-2x}\)
Evaluate from 0 to 1:
At \(x = 1\):
\(-\frac{1}{2}(2-1)e^{-2} - \frac{1}{4}e^{-2} = -\frac{1}{2}e^{-2} - \frac{1}{4}e^{-2} = -\frac{3}{4}e^{-2}\)
At \(x = 0\):
\(-\frac{1}{2}(2-0)e^{0} - \frac{1}{4}e^{0} = -1 - \frac{1}{4} = -\frac{5}{4}\)
Subtract the results:
\(\left(-\frac{3}{4}e^{-2}\right) - \left(-\frac{5}{4}\right) = \frac{5}{4} - \frac{3}{4}e^{-2}\)
Thus, the exact value is \(\frac{1}{4}(3 - e^{-2})\).
