Figure 1 shows a sketch of part of the curve with equation $y = \frac{6}{e^{x} + 2}, \; x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 5

The trapezium rule formula is given by:

$A \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_n)$

where $h$ is the width of the intervals. Here, $h = 0.2$ and the values of $y$ are:

• $y_0 = 2$
• $y_1 = 1.71830$
• $y_2 = 1.56981$
• $y_3 = 1.41994$
• $y_4 = 1.27165$
• $y_5 = 1.27165$

Calculating the area:

$A \approx \frac{0.2}{2} (2 + 2(1.71830 + 1.56981 + 1.41994 + 1.27165) + 1.27165)$

Evaluating this, we find:

$A \approx 0.2 \times 10.9999175 = 2.1999835$

Thus, rounding to four decimal places, the estimate for the area of $R$ is approximately $2.2000$.

Using the substitution $u = e^{x}$, we have:

$du = e^{x} \, dx$

Thus,

$dx = \frac{du}{u}$

The limits change accordingly:

• When $x = 0$, $u = e^{0} = 1$
• When $x = 1$, $u = e^{1} = e$

Substituting into the integral, we obtain:

$\text{Area } R = \int_{1}^{e} \frac{6}{u(u+2)} \, \frac{du}{u} = \int_{1}^{e} \frac{6}{u^2 + 2u} \, du$

This confirms that the correct expression for the area of $R$ is given by the integral with limits of $1$ to $e$.

To find the area using the derived integral, we simplify and integrate:

$\int_{1}^{e} \frac{6}{u(u+2)} \, du$

Using partial fraction decomposition:

$\frac{6}{u(u+2)} = \frac{A}{u} + \frac{B}{u+2}$

Solving for $A$ and $B$, we get:

$A(u + 2) + Bu = 6$

Setting $u = 0$ gives $A = 3$. Setting $u = -2$ gives $B = -3$. Thus:

$\int_{1}^{e} \left( \frac{3}{u} - \frac{3}{u+2} \right) du$

Evaluating the integral:

$= 3 [\ln |u|]_{1}^{e} - 3 [\ln |u+2|]_{1}^{e}$

Substituting the limits results in:

$= 3 (\ln e - \ln 1) - 3 (\ln 3 - \ln 3) = 3(1 - 0) - 3(\ln 3 - \ln 3) = 3 - 0 = 3$

Therefore, the exact area of $R$ is $3$.