\( 1< \) Find the average value of \( f(x)=-\cos x \) over the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Draw a graph of the function and indicate the average value.

To find the average value of the function \( f(x) = -\cos x \) over the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we use the formula for the average value of a function, which is: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \] Here, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \). So, we need to calculate: \[ \text{Average value} = \frac{1}{\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x \, dx \] This simplifies to: \[ = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x \, dx \] Calculating the integral: \[ \int -\cos x \, dx = -\sin x + C \] Now evaluate the definite integral: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x \, dx = \left[-\sin x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left[-\sin\left(\frac{\pi}{2}\right) - (-\sin\left(-\frac{\pi}{2}\right))\right] \] Since \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \sin\left(-\frac{\pi}{2}\right) = -1 \): \[ = [-1 - (1)] = -2 \] Thus, the integral evaluates to -2. Now plug this back into the formula for the average value: \[ \text{Average value} = \frac{1}{\pi} (-2) = -\frac{2}{\pi} \] So, the average value of \( f(x) = -\cos x \) over the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) is \( -\frac{2}{\pi} \). For graphing, you would plot the function \( f(x) = -\cos x \) over the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). The average value line would be a horizontal line at \( y = -\frac{2}{\pi} \) across this interval. The function dips down to touch -1 at \( \pm \frac{\pi}{2} \), creating a wave-like structure above this average line. Thus, the average value is represented as a horizontal line that clearly indicates where the mean value lies in relation to the oscillating cosine curve.