求 $\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl((2x-3)e^{2x}\bigr)$。 由此计算 $\displaystyle\int_0^1 4xe^{2x}\,\mathrm{d}x$ 的值，结果用精确形式表示。

1. Apply the product rule to differentiate the expression

Let $u = 2x-3$ and $v = e^{2x}$.
$\dfrac{\mathrm{d}}{\mathrm{d}x}(uv) = u\dfrac{\mathrm{d}v}{\mathrm{d}x} + v\dfrac{\mathrm{d}u}{\mathrm{d}x}$
$\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl((2x-3)e^{2x}\bigr) = (2x-3)\dfrac{\mathrm{d}}{\mathrm{d}x}(e^{2x}) + e^{2x}\dfrac{\mathrm{d}}{\mathrm{d}x}(2x-3)$

2. Differentiate the individual terms

$\dfrac{\mathrm{d}}{\mathrm{d}x}(e^{2x}) = 2e^{2x}$
$\dfrac{\mathrm{d}}{\mathrm{d}x}(2x-3) = 2$
So, the expression becomes $(2x-3)(2e^{2x}) + e^{2x}(2)$

3. Simplify the resulting expression

$= 2e^{2x}(2x-3) + 2e^{2x}$
$= e^{2x}(4x - 6 + 2)$
$= e^{2x}(4x - 4)$
$= 4(x-1)e^{2x}$

1. Find the derivative from part (a) to establish the relationship

$\frac{\mathrm{d}}{\mathrm{d}x}\bigl((2x-3)e^{2x}\bigr) = 2e^{2x} + (2x-3)(2e^{2x}) = 2e^{2x} + 4xe^{2x} - 6e^{2x} = 4xe^{2x} - 4e^{2x}$

2. Rearrange to make the required integrand the subject and integrate both sides

$4xe^{2x} = \frac{\mathrm{d}}{\mathrm{d}x}\bigl((2x-3)e^{2x}\bigr) + 4e^{2x}$

$\int 4xe^{2x}\,\mathrm{d}x = (2x-3)e^{2x} + \int 4e^{2x}\,\mathrm{d}x$
$\int 4xe^{2x}\,\mathrm{d}x = (2x-3)e^{2x} + 2e^{2x} = (2x-1)e^{2x}$

3. Evaluate the definite integral using the limits 0 and 1

$\displaystyle\int_0^1 4xe^{2x}\,\mathrm{d}x = \left[ (2x-1)e^{2x} \right]_0^1$
$= (2(1)-1)e^{2(1)} - (2(0)-1)e^{2(0)}$
$= e^2 - (-1)$
$= e^2 + 1$