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

**Step 1. Write the formula for the average value** The average value \( A \) of a function \( f(x) \) over an interval \(\left[a,b\right]\) is given by \[ A = \frac{1}{b-a}\int_{a}^{b} f(x)\,dx. \] **Step 2. Substitute the given function and interval** For \( f(x) = -\cos x \) over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we have \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \). Thus, \[ A = \frac{1}{\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x\,dx. \] Since \[ b - a = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi, \] the expression simplifies to \[ A = \frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x\,dx. \] **Step 3. Evaluate the integral** First, factor out the negative sign: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x\,dx = -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\,dx. \] We know that \[ \int \cos x\,dx = \sin x. \] Evaluating the integral, we get: \[ -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\,dx = -\Big[\sin x\Big]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = -\left(\sin \frac{\pi}{2} - \sin\left(-\frac{\pi}{2}\right)\right). \] Since \[ \sin \frac{\pi}{2} = 1 \quad \text{and} \quad \sin\left(-\frac{\pi}{2}\right) = -1, \] we have: \[ -\left(1 - (-1)\right) = -\left(1 + 1\right) = -2. \] **Step 4. Calculate the average value** Substitute the result of the integral into the average value formula: \[ A = \frac{1}{\pi} \times (-2) = -\frac{2}{\pi}. \] Thus, the average value of \( f(x) = -\cos x \) over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is \[ -\frac{2}{\pi}. \] **Graph of \( f(x) = -\cos x \) over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)** Key points on the graph: - At \( x = -\frac{\pi}{2} \): \(\cos\left(-\frac{\pi}{2}\right) = 0 \) so \( f\left(-\frac{\pi}{2}\right) = 0 \). - At \( x = 0 \): \(\cos 0 = 1 \) so \( f(0) = -1 \). - At \( x = \frac{\pi}{2} \): \(\cos\left(\frac{\pi}{2}\right) = 0 \) so \( f\left(\frac{\pi}{2}\right) = 0 \). A rough sketch of the graph: ``` f(x) | 0 | ● ● | | | -0.5 | / \ | / \ -1 | ● -1 ● | / \ |__/_____________________\________ x -π/2 0 π/2 ``` The graph is an inverted cosine curve with a minimum of \(-1\) at \( x = 0 \) and zeros at \( x = \pm \frac{\pi}{2} \).