Find the average ordinate of the sinusoid `y= sin x` over the interval `[0,pi]`

Answer

To find the average ordinate of the sinusoid \( y = \sin x \) over the interval \([0, \pi]\), we can use the formula for the average value of a function over an interval \([a, b]\):

\[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx \]

In this case, \( f(x) = \sin x \), \( a = 0 \), and \( b = \pi \).

### Step 1: Set up the integral for the average value
We need to calculate:

\[ \text{Average value} = \frac{1}{\pi - 0} \int_0^\pi \sin x \, dx \]

### Step 2: Calculate the integral \(\int_0^\pi \sin x \, dx\)
The integral of \( \sin x \) is:

\[ \int \sin x \, dx = -\cos x + C \]

Thus, we evaluate the definite integral:

\[ \int_0^\pi \sin x \, dx = \left[-\cos x\right]_0^\pi \]

Calculating this gives:

\[ = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 3: Substitute the integral back into the average value formula
Now we substitute the value of the integral back into the average value formula:

\[ \text{Average value} = \frac{1}{\pi} \cdot 2 = \frac{2}{\pi} \]

### Conclusion
Thus, the average ordinate of the sinusoid \( y = \sin x \) over the interval \([0, \pi]\) is:

\[ \frac{2}{\pi} \]

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