Problem Statement
Integration by parts formula: $\int u \frac{dv}{dx}\,dx = uv - \int v \frac{du}{dx}\,dx$.
**(a)** [7 marks] Evaluate the definite integral in exact analytical form:
$$I = \int_{0}^{1} x^2 e^{2x} \, dx$$
Verified Solution & Marking Scheme
Part (a) [M1]
First application of integration by parts: Let $u = x^2 \implies u' = 2x$, and $v' = e^{2x} \implies v = \frac{1}{2}e^{2x}$
Part (a) [A1]
$I = \left[ \frac{1}{2}x^2 e^{2x} \right]_0^1 - \int_0^1 \left(\frac{1}{2}e^{2x}\right)(2x)\,dx = \left[ \frac{1}{2}x^2 e^{2x} \right]_0^1 - \int_0^1 x e^{2x}\,dx$
Part (a) [M1]
Second application of integration by parts on $\int_0^1 x e^{2x}\,dx$: Let $u = x \implies u' = 1$, and $v' = e^{2x} \implies v = \frac{1}{2}e^{2x}$
Part (a) [A1]
$\int_0^1 x e^{2x}\,dx = \left[ \frac{1}{2}x e^{2x} \right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\,dx = \left[ \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} \right]_0^1$
Part (a) [A1]
Combine parts: $I = \left[ \frac{1}{2}x^2 e^{2x} - \frac{1}{2}x e^{2x} + \frac{1}{4}e^{2x} \right]_0^1 = \left[ e^{2x}\left(\frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{4}\right) \right]_0^1$
Part (a) [M1]
Evaluate limits: At $x = 1$: $e^2\left(\frac{1}{2} - \frac{1}{2} + \frac{1}{4}\right) = \frac{1}{4}e^2$. At $x = 0$: $e^0\left(0 - 0 + \frac{1}{4}\right) = \frac{1}{4}$.
Part (a) [A1]
$I = \frac{1}{4}e^2 - \frac{1}{4} = \frac{1}{4}(e^2 - 1)$