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