Figure 3 shows a sketch of the curve with equation $$y = \frac{2 \sin 2x}{1 + \cos x}, \quad 0 \leq x < \frac{\pi}{2}$$ The finite region $R$, shown shaded in Figure 3, is bounded by the curve and the $x$-axis - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 8

The trapezium rule states:

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

where $h$ is the width of each sub-interval. In our case, $

• h = \frac{\pi/2 - 0}{4} = \frac{\pi}{8}.$$$$ Substituting the values into the trapezium rule:

$y_0 = 0, \, y_1 = 0.75308, \, y_2 = 1.17157, \, y_3 = 1.02280, \, y_4 = 1.02280$

leads to the area:

ight)$$ Calculating this gives an area of approximately $1.1504$.

By making the substitution $u = 1 + \cos x$, we have:

$du = -\sin x \, dx.$

Thus, we rewrite the integral as:

$\int \frac{2 \sin 2x}{1 + \cos x} dx = 2 \int \sin 2x \, du.$

Using the fact that $\sin 2x = 2 \sin x \cos x$, we can evaluate:

$= 4 \int \frac{du}{u} - 4 \cos x + k$ where $k$ is a constant.

The error is calculated by comparing the exact area found through integration with the trapezium rule estimate:

Using the result from part (c): $A = 4 \ln(1 + \cos(\frac{\pi}{2})) - 4 \cos(\frac{\pi}{2}) + k = 1.1504$

Thus, the error in the estimate is approximately: $\text{Error} = |4 - 1.1504| \approx 2.8496.$

Rounded to 2 significant figures, the error is approximately $2.8$.