o) Hence find \( \int x \mathrm{e}^{-2 x} \mathrm{~d} x \)

To solve the integral \( \int x e^{-2x} \, dx \), we can use integration by parts. The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] We will choose: - \( u = x \) (which means \( du = dx \)) - \( dv = e^{-2x} \, dx \) (which means we need to find \( v \)) To find \( v \), we integrate \( dv \): \[ v = \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \] Now we can apply the integration by parts formula: \[ \int x e^{-2x} \, dx = uv - \int v \, du \] Substituting in our choices for \( u \), \( du \), \( v \), and \( dv \): \[ \int x e^{-2x} \, dx = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) \, dx \] This simplifies to: \[ -\frac{1}{2} x e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \] We already computed \( \int e^{-2x} \, dx \) earlier, which is: \[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \] Substituting this back into our equation gives: \[ -\frac{1}{2} x e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) \] This simplifies to: \[ -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} \] Thus, the integral is: \[ \int x e^{-2x} \, dx = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C \] where \( C \) is the constant of integration. Therefore, the final answer is: \[ \int x e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \left( x + \frac{1}{2} \right) + C \]