Let $h(x) = ext{cos}(2x)$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2018

The average value of a function $f(x)$ over an interval $[a, b]$ is given by:
$\frac{1}{b-a} \int_a^b f(x) \ dx.$
For our case:
$\text{Average value of } h(x) = \frac{1}{\frac{\pi}{4} - 0} \int_0^{\frac{\pi}{4}} \cos(2x) \ dx.$
This simplifies to:
$\frac{4}{\pi} \int_0^{\frac{\pi}{4}} \cos(2x) \ dx.$
The integral evaluates as follows:
$\int \cos(2x) \ dx = \frac{1}{2}\sin(2x) + C.$
Now applying the limits:
$\left[\frac{1}{2}\sin(2x)\right]_0^{\frac{\pi}{4}} = \frac{1}{2} \left(\sin\left(\frac{\pi}{2}\right) - \sin(0)\right) = \frac{1}{2} \cdot 1 = \frac{1}{2}.$
Putting it all together:
$\text{Average value} = \frac{4}{\pi} \cdot \frac{1}{2} = \frac{2}{\pi}.$