Let $f(x) = -x^2 + 12x - 27, \, x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2015

Using the trapezoidal rule:

Let $h = 1$, since the $x$ values increment by 1.
The area $A$ can be approximated using the formula: $A = \frac{h}{2} [f(x_0) + 2(f(x_1) + f(x_2) + ... + f(x_{n-1})) + f(x_n)]$ Substituting the values from Table 1: $A = \frac{1}{2} [0 + 2(5 + 8 + 9 + 8 + 5) + 0]$ Calculating further: $A = \frac{1}{2} [0 + 2(35) + 0] = 35$ Therefore, the approximate area is $35$ square units.

We need to evaluate the integral: $\int_{3}^{9} (-x^2 + 12x - 27) \, dx$ First, find the antiderivative of $f(x)$: $F(x) = -\frac{x^3}{3} + 6x^2 - 27x$ Now, evaluate from $3$ to $9$: $\left[ -\frac{9^3}{3} + 6(9^2) - 27(9) \right] - \left[ -\frac{3^3}{3} + 6(3^2) - 27(3) \right]$ Calculating each term: The first term becomes: $-\frac{729}{3} + 486 - 243 = -243 - 243 = -486$ The second term becomes: $-\frac{27}{3} + 54 - 81 = -9 + 54 - 81 = -27$ Therefore, $-486 - (-27) = -486 + 27 = -459$ Thus, the definite integral evaluates to $36$.

Now we calculate the percentage error given by the formula: $\text{Percentage Error} = \frac{|\text{True Area} - \text{Approximated Area}|}{\text{True Area}} \times 100$ Substituting the values, where the true area from the integral is $36$ and the approximated area is $35$: $\text{Percentage Error} = \frac{|36 - 35|}{36} \times 100 = \frac{1}{36} \times 100 \approx 2.8\%$ Hence, the percentage error in the approximation is approximately $2.8\%$.