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

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To find the average value of the function $$ f(t) = \sin(t) + \cos(t) $$ over the interval $$[0, \pi]$$, we use the formula for the average value of a function over an interval $$[a, b]$$:

$$
\text{Average value} = \frac{1}{b - a} \int_a^b f(t) \, dt
$$

Plugging in the given function and interval:

$$
\text{Average value} = \frac{1}{\pi - 0} \int_0^\pi (\sin(t) + \cos(t)) \, dt = \frac{1}{\pi} \left( \int_0^\pi \sin(t) \, dt + \int_0^\pi \cos(t) \, dt \right)
$$

Now, compute each integral separately:

1. **Integral of $$\sin(t)$$:**
$$
\int_0^\pi \sin(t) \, dt = \left[ -\cos(t) \right]_0^\pi = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2
$$

2. **Integral of $$\cos(t)$$:**
$$
\int_0^\pi \cos(t) \, dt = \left[ \sin(t) \right]_0^\pi = \sin(\pi) - \sin(0) = 0 - 0 = 0
$$

Adding these results:

$$
\int_0^\pi (\sin(t) + \cos(t)) \, dt = 2 + 0 = 2
$$

Finally, compute the average value:

$$
\text{Average value} = \frac{2}{\pi}
$$

**Answer:**  
Its average value over $$[0, \pi]$$ is $$ \dfrac{2}{\,\pi\,} $$.
The average value of $$ f(t) = \sin(t) + \cos(t) $$ over $$[0, \pi]$$ is $$ \frac{2}{\pi} $$.

Compute the average value of over the interval .

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