The height of an arch above the ground is given by the function \( y=8 \cos x \) for \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). What is the average height of the arch above the ground? The average height is (Type an exact answer, using \( \pi \) as needed.)

To find the average height of the arch given by the function \( y = 8 \cos x \) over the interval \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \), we use the formula for the average value of a function: \[ \text{Average height} = \frac{1}{b-a} \int_a^b f(x) \, dx \] In this case, \( a = -\frac{\pi}{2} \), \( b = \frac{\pi}{2} \), and \( f(x) = 8 \cos x \). Calculating the integral: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 8 \cos x \, dx \] The integral of \( 8 \cos x \) is \( 8 \sin x \), so we evaluate: \[ = 8 \left[ \sin x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 8 \left( \sin \frac{\pi}{2} - \sin \left(-\frac{\pi}{2}\right) \right) \] \[ = 8 \left(1 - (-1)\right) = 8 \cdot 2 = 16 \] Now, substituting into the formula for the average height: \[ \text{Average height} = \frac{1}{\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)} \cdot 16 = \frac{1}{\pi} \cdot 16 = \frac{16}{\pi} \] Thus, the average height of the arch above the ground is: \[ \frac{16}{\pi} \]