Problem Statement
Consider $f(x) = x^2 e^{-2x}$.
(a) Find $\int x e^{-2x} dx$. [3 marks]
(b) Hence, evaluate $\int_{0}^{1} x^2 e^{-2x} dx$ in the form $\frac{a - b e^{-2}}{c}$. [4 marks]
Verified Solution & Marking Scheme
Part (a): Parts on x e⁻²ˣ
$u = x, dv = e^{-2x}dx \implies -\frac{1}{2}x e^{-2x} + \frac{1}{2}\int e^{-2x}dx = -\frac{1}{4}(2x + 1)e^{-2x} + C$
Part (b): Parts on x² e⁻²ˣ
$\int_{0}^{1} x^2 e^{-2x} dx = \left[-\frac{1}{2}x^2 e^{-2x}\right]_{0}^{1} + \int_{0}^{1} x e^{-2x} dx = \left[-\frac{1}{4}(2x^2 + 2x + 1)e^{-2x}\right]_{0}^{1}$
Evaluate Limits
$-\frac{5}{4}e^{-2} - \left(-\frac{1}{4}\right) = \frac{1 - 5e^{-2}}{4}$