Compute the average value of \( f(t) = \sin(t) + \cos(t) \) over the interval \([0, \pi]\).

To find the average value of the function \( f(t) = \sin(t) + \cos(t) \) over the interval \([0, \pi]\), we can use the formula for the average value of a function over an interval \([a, b]\): \[ \text{Average value} = \frac{1}{b - a} \int_a^b f(t) \, dt \] In this case, \( a = 0 \) and \( b = \pi \). First, we compute the definite integral of \( f(t) \): \[ \int_0^\pi (\sin(t) + \cos(t)) \, dt \] This can be split into two integrals: \[ \int_0^\pi \sin(t) \, dt + \int_0^\pi \cos(t) \, dt \] Now, we calculate each integral separately. 1. **Calculating \(\int_0^\pi \sin(t) \, dt\)**: \[ \int \sin(t) \, dt = -\cos(t) + C \] Evaluating from \(0\) to \(\pi\): \[ \left[-\cos(t)\right]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \] 2. **Calculating \(\int_0^\pi \cos(t) \, dt\)**: \[ \int \cos(t) \, dt = \sin(t) + C \] Evaluating from \(0\) to \(\pi\): \[ \left[\sin(t)\right]_0^\pi = \sin(\pi) - \sin(0) = 0 - 0 = 0 \] Now we combine the results from both integrals: \[ \int_0^\pi (\sin(t) + \cos(t)) \, dt = 2 + 0 = 2 \] Next, we find the average value: \[ \text{Average value} = \frac{1}{\pi - 0} \int_0^\pi f(t) \, dt = \frac{1}{\pi} \cdot 2 = \frac{2}{\pi} \] Thus, the average value of \( f(t) = \sin(t) + \cos(t) \) over the interval \([0, \pi]\) is \[ \boxed{\frac{2}{\pi}} \]