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

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UpStudy AI · Accepted Answer

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 $$[a, b]$$:

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

Here, $$ a = 0 $$ and $$ b = \pi $$. Plugging in the values:

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

Let's compute each integral separately:

1. **Integral of $$ \sin(t) $$:**
   $$
   \int_{0}^{\pi} \sin(t) \, dt = [-\cos(t)]_{0}^{\pi} = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2
   $$

2. **Integral of $$ \cos(t) $$:**
   $$
   \int_{0}^{\pi} \cos(t) \, dt = [\sin(t)]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0
   $$

Adding these results:

$$
\text{Average value} = \frac{1}{\pi} (2 + 0) = \frac{2}{\pi}
$$

**Final Answer:**

$$
\boxed{\dfrac{2}{\pi}}
$$
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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