Figure 1 shows a sketch of part of the curve with equation y = (2 - x)e^{2x}, ext{x} ext{ in } ext{R} The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the y-axis - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 8

The trapezium rule can be made more accurate by:

• Increasing the number of intervals (or strips) used, which leads to a finer approximation.
• Making the width of the trapezoids smaller, resulting in better approximations of the curve's shape.
• Utilizing more values of x (more x-coordinates) to gather more sample points of the function.

To find the exact area under the curve, we set up the integral:

$A = \int_0^2 (2 - x)e^{2x} \, dx$

We can use integration by parts: Let:

• $u = (2 - x)$, so that $du = -dx$
• $dv = e^{2x} dx$, hence $v = \frac{1}{2}e^{2x}$

Applying integration by parts gives:

$A = \left[ \frac{1}{2}(2 - x)e^{2x} \right]_0^2 - \int_0^2 -\frac{1}{2}e^{2x} \, dx$

Calculating the first term:
At x=2, $(2 - 2)e^{4} = 0$ At x=0, $(2 - 0)e^{0} = 2$. Therefore,
$A = \frac{1}{2}(0 - 2) - \int_0^2 -\frac{1}{2}e^{2x} \, dx$

Evaluating the integral: $\int -\frac{1}{2}e^{2x} \, dx = -\frac{1}{4}e^{2x}$

Thus, $A = -1 + \left(-\frac{1}{4}e^{4} + \frac{1}{4}e^{0} \right) = -1 - \frac{1}{4}(e^{4} - 1)$

To finalize the area: $A = \frac{1}{4} - 1 - \frac{1}{4}e^{4}$

Hence, the exact area is: $\frac{1}{4} - 1 - \frac{1}{4}e^{4} = -3.25 - \frac{1}{4}e^{4}$