Find the average value of the following function over the given interval. Draw a graph of the \( f(x)=\frac{12}{x^{2}+1} ;[-1,1] \)

**Step 1. Write the formula for the average value of a function** The average value \( A \) of a function \( f(x) \) over an interval \([a,b]\) is given by \[ A = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] For the function \( f(x)=\frac{12}{x^2+1} \) over the interval \([-1, 1]\), we have \[ A = \frac{1}{1-(-1)} \int_{-1}^{1} \frac{12}{x^2+1} \, dx = \frac{1}{2} \int_{-1}^{1} \frac{12}{x^2+1} \, dx. \] --- **Step 2. Evaluate the integral** Factor out the constant: \[ \int_{-1}^{1} \frac{12}{x^2+1} \, dx = 12 \int_{-1}^{1} \frac{1}{x^2+1} \, dx. \] Recall that \[ \int \frac{1}{x^2+1} \, dx = \arctan x + C. \] Thus, \[ \int_{-1}^{1} \frac{1}{x^2+1} \, dx = \arctan(1) - \arctan(-1). \] Since \[ \arctan(1) = \frac{\pi}{4} \quad \text{and} \quad \arctan(-1) = -\frac{\pi}{4}, \] we have \[ \arctan(1) - \arctan(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}. \] Thus, the integral becomes \[ 12 \cdot \frac{\pi}{2} = 6\pi. \] --- **Step 3. Compute the average value** Substitute the value of the integral into the average value formula: \[ A = \frac{1}{2} \cdot 6\pi = 3\pi. \] So, the average value of \( f(x) \) over \([-1,1]\) is \[ \boxed{3\pi}. \] --- **Step 4. Graph the function \( f(x)=\frac{12}{x^2+1} \) on \([-1,1]\)** - **Symmetry:** The function is even since it depends on \( x^2 \). Its graph is symmetric about the \( y \)-axis. - **Key Points:** - At \( x=0 \): \[ f(0) = \frac{12}{0^2+1} = 12. \] - At \( x=1 \) and \( x=-1 \): \[ f(1) = f(-1) = \frac{12}{1+1} = 6. \] - **Shape:** The graph is a smooth curve that attains its maximum value at \( x=0 \) and decreases symmetrically toward the endpoints at \( x = -1 \) and \( x = 1 \). A rough sketch of the graph is as follows: ``` f(x) │ 12 ─● ● │ \ │ \ │ \ 9 │ \ │ \ │ ● 6 ─●──────────────●───────────── x │ -1 0 1 ``` - The point \( (0, 12) \) is the maximum. - The points \( (-1, 6) \) and \( (1, 6) \) are the endpoints on the interval. - The curve is smooth and symmetric about the \( y \)-axis. --- **Final Answer** The average value of \( f(x)=\frac{12}{x^2+1} \) over \([-1,1]\) is \[ \boxed{3\pi}. \]