\( \left(1.0\right. \) punto) \( \int_{1}^{2} 2 x e^{-3 x} d x \)

To solve the integral \( \int_{1}^{2} 2 x e^{-3 x} d x \), we can use integration by parts. Set \( u = 2x \) and \( dv = e^{-3x} dx \). Then, calculate \( du = 2 dx \) and \( v = -\frac{1}{3} e^{-3x} \). Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get: \[ \int 2x e^{-3x} dx = -\frac{2}{3} x e^{-3x} + \frac{2}{9} e^{-3x} + C \] Now, evaluate this from \( x = 1 \) to \( x = 2 \): \[ \left[-\frac{2}{3} \cdot 2 e^{-6} + \frac{2}{9} e^{-6}\right] - \left[-\frac{2}{3} \cdot 1 e^{-3} + \frac{2}{9} e^{-3}\right] \] Calculating those values gives you the definite integral result effectively!