Figure 1 shows a sketch of part of the curve with equation $y = \frac{6}{e^{x} + 2}, \; x \in \mathbb{R}$ 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 3 - 2017 - Paper 5

Using the trapezium rule, the area $A$ can be estimated as:

$A \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_5)$

where $h = 0.2$, and the $y$ values are:

• $y_0 = 2$
• $y_1 = 1.71830$
• $y_2 = 1.56981$
• $y_3 = 1.41994$
• $y_4 = 1.27165$
• $y_5 = 1.20000$

Calculating:

$A \approx \frac{0.2}{2} (2 + 2(1.71830) + 2(1.56981) + 2(1.41994) + 2(1.27165) + 1.20000)$

Evaluating this gives:

$A \approx 0.1 (2 + 3.43660 + 3.13962 + 2.83988 + 2.54330 + 1.20000) \approx 0.1 (15.15940) \approx 1.51594$

Thus, the estimated area $A \approx 1.5159$.

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

• Then, $du = e^{x} \, dx = u \, dx \implies dx = \frac{du}{u}$.

• The limits change as follows:

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

Now substituting into the integral for the area:

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

And so this shows that the area can be represented as:

$\int_{0}^{6} \frac{6}{u(u + 2)} \, du$

To find the exact area, we evaluate the integral:

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

Using partial fractions, we write:

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

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

• $A = 3$, $B = -3$

Thus:

$A = \int \left( \frac{3}{u} - \frac{3}{u + 2} \right) du = 3 \ln|u| - 3 \ln|u + 2| + C$

Now applying the limits:

$= \left( 3 \ln(e) - 3 \ln(e + 2) \right) - \left( 3 \ln(1) - 3 \ln(3) \right)$

Evaluating yields:

$= 3 - 3 \ln(3) + 3 \ln(4)$

Thus, the exact area of $R$ is:

$A \approx 3 \ln \left( \frac{4}{3} \right).$