Use the function $h(x) = 0.001x^3 - 0.12x^2 + 3.6x + 5$, $x \in \mathbb{R}$, to find the average height of this section of the track above level ground, from $x = 0$ to $x = 75$ - Leaving Cert Mathematics - Question c - 2021

Next, we evaluate the definite integral from 0 to 75:

$\int_0^{75} h(x) \; dx = \left[ \frac{0.001(75)^4}{4} - \frac{0.12(75)^3}{3} + \frac{3.6(75)^2}{2} + 5(75) \right] - \left[ 0 \right]$

Calculating this step-by-step, we find:

• For $x = 75$:
  • $\frac{0.001(75)^4}{4} = 1535.15625$
  • $\frac{0.12(75)^3}{3} = 156.25$
  • $\frac{3.6(75)^2}{2} = 1350$
  • $5(75) = 375$

Thus, the total area under $h(x)$ from $0$ to $75$ is:

$1535.15625 - 156.25 + 1350 + 375 = 1547.90625$

Now, to find the average height, we divide the total area by the length of the interval:

$\text{Average height} = \frac{1}{75} \int_0^{75} h(x) \; dx = \frac{1547.90625}{75} \approx 20.639\text{ m}$

Rounding to two decimal places, we get:

$\text{Average height} \approx 20.64 \text{ m}$