回答所有问题。

1. Apply the product rule to differentiate the expression

Let $y = 2xe^{1-2x}$.
Using the product rule $\frac{\text{d}}{\text{d}x}(uv) = u\frac{\text{d}v}{\text{d}x} + v\frac{\text{d}u}{\text{d}x}$ with $u = 2x$ and $v = e^{1-2x}$:
$\frac{\text{d}y}{\text{d}x} = 2x \cdot \frac{\text{d}}{\text{d}x}(e^{1-2x}) + e^{1-2x} \cdot \frac{\text{d}}{\text{d}x}(2x)$

2. Differentiate the exponential term using the chain rule and simplify

$\frac{\text{d}y}{\text{d}x} = 2x(-2e^{1-2x}) + e^{1-2x}(2)$
$= -4xe^{1-2x} + 2e^{1-2x}$
$= 2e^{1-2x}(1 - 2x)$

1. Find the derivative of $2xe^{1-2x}$ from part (i) to use for integration.

Let $y = 2xe^{1-2x}$. Using the product rule:
$\frac{\text{d}y}{\text{d}x} = 2(e^{1-2x}) + 2x(-2e^{1-2x})$
$\frac{\text{d}y}{\text{d}x} = 2e^{1-2x} - 4xe^{1-2x}$

2. Rearrange the derivative to make $2xe^{1-2x}$ the subject.

$4xe^{1-2x} = 2e^{1-2x} - \frac{\text{d}}{\text{d}x}(2xe^{1-2x})$
Dividing by 2:
$2xe^{1-2x} = e^{1-2x} - \frac{1}{2}\frac{\text{d}}{\text{d}x}(2xe^{1-2x})$

3. Integrate both sides with respect to $x$ from 0 to 1.

$\int_{0}^{1} 2xe^{1-2x} \text{d}x = \int_{0}^{1} e^{1-2x} \text{d}x - \frac{1}{2} \int_{0}^{1} \frac{\text{d}}{\text{d}x}(2xe^{1-2x}) \text{d}x$
$\int_{0}^{1} 2xe^{1-2x} \text{d}x = \left[ \frac{e^{1-2x}}{-2} \right]_{0}^{1} - \frac{1}{2} \left[ 2xe^{1-2x} \right]_{0}^{1}$

4. Evaluate the definite integral to find the exact value.

$= \left( \frac{e^{-1}}{-2} - \frac{e^{1}}{-2} \right) - \frac{1}{2} \left( 2(1)e^{-1} - 0 \right)$
$= -\frac{1}{2e} + \frac{e}{2} - \frac{1}{e}$
$= \frac{e}{2} - \frac{3}{2e}$

1. Determine the sign of the function for positive x to distinguish between the graphs

For $x > 0$, $2x > 0$ and $e^{1-2x} > 0$, which means $y = 2xe^{1-2x} > 0$.
Alternatively, using the derivative from (i), $\frac{\text{d}y}{\text{d}x} = 2e^{1-2x}(1 - 2x)$. At $x = 0$, the gradient is $2e > 0$, so the graph is initially increasing.

2. Conclude which graph matches the deduced properties

Graph 1 is positive for $x > 0$ and has a positive initial gradient, whereas Graph 2 dips below the $x$-axis and has a negative initial gradient. Therefore, Graph 1 is the correct graph.