Figure 1 shows a sketch of part of the curve with equation $y = \frac{6}{e^{x} + 2}, \quad x \in \mathbb{R}$ The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $y$-axis, the $x$-axis and the line with equation $x = 1$ - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 5

Let the trapezium rule be applied on the interval $[0, 1]$ using the known values:

$A \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_5),$
where $h = (1 - 0) / 5 = 0.2$ and the values of $y$ are:

• $y_0 = 2$
• $y_1 = 1.86054$
• $y_2 = 1.71830$
• $y_3 = 1.56981$
• $y_4 = 1.41994$
• $y_5 = 1.27165$

Hence, computing the areas gives:

$A \approx \frac{0.2}{2} (2 + 2(1.86054) + 2(1.71830) + 2(1.56981) + 2(1.41994) + 1.27165) \approx 0.1(2 + 3.72108 + 3.43660 + 3.13962 + 2.83988 + 1.27165)\approx 0.1(16.40843) \approx 1.640843.$

Thus, the estimated area of $R$ is approximately $1.6408$.

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

$du = e^{x} dx = u dx \implies dx = \frac{du}{u}.$
Given the bounds for $x$: when $x = 0$, $u = e^{0} = 1$, and when $x = 1$, $u = e^{1} = e$. We rewrite the integral as:

$A = \int_{0}^{1} \frac{6}{e^{x} + 2} \, dx = \int_{1}^{e} \frac{6}{u + 2} \cdot \frac{du}{u} = \int_{1}^{e} \frac{6}{u(u + 2)} \, du.$
Thus, confirming the area of $R$ can be expressed as stated.

To calculate the area of $R$, we compute:

$A = \int_{1}^{e} \frac{6}{u(u + 2)} \, du.$
Using partial fractions, we can write:

$\frac{6}{u(u + 2)} = \frac{A}{u} + \frac{B}{u + 2}$
To find $A$ and $B$, we multiply both sides by $u(u + 2)$, resulting in:

$6 = A(u + 2) + Bu.$
Setting $u = 0$, we find:

$6 = 2A \implies A = 3.$
Setting $u = -2$, we find:

$6 = -2B \implies B = -3.$
So we have:

$A = \int_{1}^{e} \left( \frac{3}{u} - \frac{3}{u + 2} \right) \, du = 3\ln |u| - 3\ln |u + 2| \bigg|_{1}^{e}.$
Using the limits:

$= 3(\ln e - \ln 3) - 3(\ln 1 - \ln 3) = 3(1 - \ln 3) - 3(-\ln 3) = 3(1) = 3.$
Thus, the exact area of $R$ is $3$.