Problem Statement
Consider the integral $I_n = \int_{0}^{1} x^n e^{2x} \, dx$, where $n \in \mathbb{Z}^+$.
(a) Show that $I_n = \frac{e^2}{2} - \frac{n}{2} I_{n-1}$ for $n \ge 1$. [4 marks]
(b) Hence, evaluate $I_2$ in exact form. [3 marks]
Verified Solution & Marking Scheme
Part (a): Integration by parts setup
Let $u = x^n \implies du = n x^{n-1} \, dx$.
Let $dv = e^{2x} \, dx \implies v = \frac{1}{2} e^{2x}$.
Using $\int u \, dv = uv - \int v \, du$:
$$I_n = \left[ \frac{1}{2} x^n e^{2x} \right]_{0}^{1} - \int_{0}^{1} \frac{1}{2} e^{2x} (n x^{n-1}) \, dx$$
$$I_n = \left( \frac{1}{2}(1)^n e^2 - 0 \right) - \frac{n}{2} \int_{0}^{1} x^{n-1} e^{2x} \, dx = \frac{e^2}{2} - \frac{n}{2} I_{n-1} \quad \text{(AG)}$$
Part (b): Evaluate base case I_0
$$I_0 = \int_{0}^{1} e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_{0}^{1} = \frac{e^2 - 1}{2}$$
Compute I_1 and then I_2
Using recurrence relation:
$$I_1 = \frac{e^2}{2} - \frac{1}{2} I_0 = \frac{e^2}{2} - \frac{1}{2}\left(\frac{e^2 - 1}{2}\right) = \frac{e^2}{2} - \frac{e^2 - 1}{4} = \frac{e^2 + 1}{4}$$
Now compute $I_2$:
$$I_2 = \frac{e^2}{2} - \frac{2}{2} I_1 = \frac{e^2}{2} - I_1 = \frac{e^2}{2} - \frac{e^2 + 1}{4} = \frac{2e^2 - (e^2 + 1)}{4} = \frac{e^2 - 1}{4}$$