Let's solve the problem step-by-step. **Given:** \[ g(t) = \frac{3}{1 + t^2}, \quad \text{interval: } [0, 4] \] The **average value** of a function \(g(t)\) on the interval \([a, b]\) is given by: \[ g_{\text{ave}} = \frac{1}{b-a} \int_a^b g(t)\,dt \] Here, \(a = 0\), \(b = 4\). --- ### 1. Write the formula for the average value: \[ g_{\text{ave}} = \frac{1}{4-0} \int_0^4 \frac{3}{1 + t^2}\, dt \] \[ g_{\text{ave}} = \frac{1}{4} \int_0^4 \frac{3}{1 + t^2}\, dt \] --- ### 2. Factor out the constant: \[ g_{\text{ave}} = \frac{3}{4} \int_0^4 \frac{1}{1 + t^2}\, dt \] --- ### 3. Integrate Recall: \[ \int \frac{1}{1 + t^2}\, dt = \arctan(t) \] So: \[ g_{\text{ave}} = \frac{3}{4} \left[\arctan(t)\right]_0^4 \] --- ### 4. Evaluate the definite integral: \[ g_{\text{ave}} = \frac{3}{4} \left(\arctan(4) - \arctan(0)\right) \] \[ \arctan(0) = 0 \] So: \[ g_{\text{ave}} = \frac{3}{4} \arctan(4) \] --- ### 5. Final answer (exact and decimal): **Exact form:** \[ \boxed{g_{\text{ave}} = \frac{3}{4} \arctan(4)} \] **Decimal value:** \[ \arctan(4) \approx 1.3258 \] \[ g_{\text{ave}} \approx \frac{3}{4} \times 1.3258 \approx 0.9944 \] **Final answer:** \[ \boxed{g_{\text{ave}} \approx 0.994} \] Let me know if you need further clarification!