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 above the ground given by the function \( y = 8 \cos x \) over the interval \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \), we can use the formula for the average value of a function over an interval \([a, b]\): \[ \text{Average height} = \frac{1}{b - a} \int_a^b f(x) \, dx \] In this case, \( f(x) = 8 \cos x \), \( a = -\frac{\pi}{2} \), and \( b = \frac{\pi}{2} \). 1. Calculate the length of the interval: \[ b - a = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \] 2. Set up the integral: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 8 \cos x \, dx \] 3. Calculate the integral: \[ \int 8 \cos x \, dx = 8 \sin x \] 4. Evaluate the definite integral from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \): \[ \left[ 8 \sin x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 8 \sin\left(\frac{\pi}{2}\right) - 8 \sin\left(-\frac{\pi}{2}\right) \] \[ = 8(1) - 8(-1) = 8 + 8 = 16 \] 5. Now, substitute back into the average height formula: \[ \text{Average height} = \frac{1}{\pi} \cdot 16 = \frac{16}{\pi} \] Thus, the average height of the arch above the ground is \[ \frac{16}{\pi} \]