Figure 1 shows a sketch of part of the curve with equation $y = (2 - x)e^{2x}$, $x \in \mathbb{R}$ 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

To improve the accuracy of the approximation using the trapezium rule, one can:

• Increase the number of strips: Using more subdivisions leads to a closer approximation to the actual curve.
• Make h smaller: Decreasing the width of each interval reduces the approximation error.
• Use more values of $x$: By calculating more $y$ values, we can better capture the shape of the curve.

To find the exact area of the region $R$, we need to integrate the function from $x = 0$ to $x = 2$:

1. Set up the integral: $A = \int_0^2 (2 - x)e^{2x} \, dx$

2. To solve the integral, we can use integration by parts, where:

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

3. Applying integration by parts: $\int u \, dv = uv - \int v \, du$ $= (2 - x) \cdot \frac{1}{2} e^{2x} \bigg|_0^2 + \int \frac{1}{2} e^{2x} \, dx$

4. Evaluating the boundary terms: At $x = 2$, $u = 0$, and at $x = 0$, $u = 2$: $= [(2 - 2) \cdot \frac{1}{2} e^{4}] - [(2 - 0) \cdot \frac{1}{2} e^0]$ $= 0 - 1 = -1$

5. Now compute the integral: $\frac{1}{2} \int e^{2x} dx = \frac{1}{2} \cdot \frac{1}{2} e^{2x} + C$ Evaluating from $0$ to $2$ gives: $= \frac{1}{4} [e^4 - 1]$

Combining the results: $A = -1 + \frac{1}{4}(e^4 - 1)$ $= \frac{1}{4} e^4 - 1 - \frac{1}{4}$ $= \frac{1}{4} e^4 - \frac{5}{4}$

The exact value of the area $R$ is: $A = \frac{1}{4}(e^4 - 5)$