Differentiate $4xe^{3x}$ with respect to $x$. Hence, evaluate $\int_{0}^{3} 6xe^{3x} \text{d}x$, giving your answer in exact form.

1. Identify the terms for the product rule and find their derivatives

Let $u = 4x$ and $v = e^{3x}$.
Then $\frac{\text{d}u}{\text{d}x} = 4$ and $\frac{\text{d}v}{\text{d}x} = 3e^{3x}$.

2. Apply the product rule and simplify

$\frac{\text{d}}{\text{d}x}(4xe^{3x}) = u\frac{\text{d}v}{\text{d}x} + v\frac{\text{d}u}{\text{d}x}$
$= (4x)(3e^{3x}) + (e^{3x})(4)$
$= 12xe^{3x} + 4e^{3x}$
$= 4e^{3x}(3x + 1)$

1. Find the derivative of $4xe^{3x}$ from part (a) to use for the 'Hence' method.

$\frac{\text{d}}{\text{d}x}(4xe^{3x}) = 4e^{3x} + 4x(3e^{3x}) = 4e^{3x} + 12xe^{3x}$

2. Use integration as the reverse of differentiation to express the integral of $12xe^{3x}$.

$\int (4e^{3x} + 12xe^{3x}) \text{d}x = 4xe^{3x}$
$\int 12xe^{3x} \text{d}x = 4xe^{3x} - \int 4e^{3x} \text{d}x$

3. Relate the result to the required integral $\int 6xe^{3x} \text{d}x$ and integrate the remaining exponential term.

$\int 6xe^{3x} \text{d}x = \frac{1}{2} \int 12xe^{3x} \text{d}x$
$= \frac{1}{2} \left( 4xe^{3x} - \int 4e^{3x} \text{d}x \right)$
$= 2xe^{3x} - \int 2e^{3x} \text{d}x$
$= 2xe^{3x} - \frac{2}{3}e^{3x}$

4. Evaluate the definite integral from $0$ to $3$.

$\int_{0}^{3} 6xe^{3x} \text{d}x = \left[ 2xe^{3x} - \frac{2}{3}e^{3x} \right]_{0}^{3}$
$= \left( 2(3)e^{3(3)} - \frac{2}{3}e^{3(3)} \right) - \left( 2(0)e^{0} - \frac{2}{3}e^{0} \right)$
$= \left( 6e^9 - \frac{2}{3}e^9 \right) - \left( 0 - \frac{2}{3} \right)$
$= \frac{16}{3}e^9 + \frac{2}{3}$