\( 1 \leftarrow \quad \) Find the average value of \( f(x)=-\cos x \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \). Draw a gra

To find the average value of the function \( f(x) = -\cos x \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \), we use the formula for the average value of a function: \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] In this case, \( a = -\frac{\pi}{4} \) and \( b = \frac{\pi}{4} \). First, calculate the length of the interval: \[ b - a = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \] Now, we compute the integral: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} -\cos x \, dx \] This simplifies to: \[ - \left[ \sin x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = -\left( \sin\left(\frac{\pi}{4}\right) - \sin\left(-\frac{\pi}{4}\right) \right) = -\left( \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right) \right) \] This becomes: \[ - \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) = -\sqrt{2} \] Now, substituting back into the average value formula: \[ \text{Average Value} = \frac{1}{\frac{\pi}{2}} (-\sqrt{2}) = -\frac{2\sqrt{2}}{\pi} \] So the average value of \( f(x) = -\cos x \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \) is \[ -\frac{2\sqrt{2}}{\pi}. \] --- To visualize this, you could graph \( f(x) = -\cos x \). The curve is shaped like an upside-down wave peaking at 0, dipping to -1 at \( x = \pm \frac{\pi}{2}, \) and having a value of -0.707 (i.e., \(-\frac{\sqrt{2}}{2}\)) at \( x = \pm \frac{\pi}{4}. \) The average value reflects the displacement of the wave over the specified interval, graphed within the bounds of the x-axis. An engaging element to think about: Did you know this average value can be interpreted as the "mean height" of that wave over the interval? It showcases how we can describe not just a value but the overall "behavior" of the function in that span!