Figure 1 shows part of the curve with equation $x = 4e^{t/3} + 3$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Using the trapezium rule, the area can be estimated using:

$A \approx \frac{h}{2} [y_0 + 2y_1 + 2y_2 + y_n]$

where $h$ is the width of each interval. Here, $h = 2$ (intervals between $t$ values: 0, 2, 4, 6, 8).

Substituting in the values:

$A \approx \frac{2}{2} [3 + 2(7.107 + 7.218) + 5.223]$

Calculating this gives:

$A \approx 1 [3 + 2(14.325) + 5.223] \approx 1 [3 + 28.650 + 5.223] \approx 36.873$

Therefore, the area is approximately 36.87 (rounded to 2 decimal places).

To find the exact area of region $R$, we need to calculate the integral:

$\int_0^8 (4e^{t/3} + 3) \, dt$

Calculating each part separately:

1. The integral of $4e^{t/3}$ is $12e^{t/3}$.
2. The integral of 3 is $3t$.

Thus, we can evaluate:

$\left[12e^{t/3} + 3t \right]_0^8$

Substituting the limits:

1. For $t = 8$: $12e^{8/3} + 24$.
2. For $t = 0$: $12e^0 + 0 = 12$.

Thus:

$Area = (12e^{8/3} + 24 - 12)$

Calculating gives us the exact area.

Let $A_b$ be the area from part (b) and $A_c$ from part (c). The difference is:

$\text{Difference} = |A_b - A_c|$

Calculate the difference using the approximate area from part (b) and the exact area computed in part (c). Round to two decimal places.