Question

The average ordinate of y=sin x over [0,\(\pi\)] is

• A. \(\frac{2}{\pi}\)
• B. \(\frac{3}{\pi}\)
• C. \(\frac{4}{\pi}\)
• D. \(\pi\)

Answer 1

The Correct Option is A

Given Information 

The correct answer is \( \frac{2}{\pi} \).

The mean value \( \mu \) is given by:

\[ \mu=\frac{1}{b-a}\int_{a}^{b} f(x)dx \]

Applying the Formula

In this specific case, we are calculating the mean value \( \mu_y \) of \( f(x) = \sin(x) \) over the interval \( [0, \pi] \). Therefore, \( a = 0 \) and \( b = \pi \).

\[ \mu_y=\frac{1}{\pi-0}\int_{0}^{\pi} \sin(x) \, dx \]

Calculating the Integral

Now, we evaluate the integral:

\[ \int_{0}^{\pi} \sin(x) \, dx = [-\cos(x)]_{0}^{\pi} \]

Evaluating the Limits

Substitute the limits of integration:

\[ [-\cos(x)]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

Finding the Mean Value

Substitute the value of the integral back into the formula for \( \mu_y \):

\[ \mu_y = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) \, dx = \frac{1}{\pi} (2) = \frac{2}{\pi} \]

Conclusion

Therefore, the mean value \( \mu_y \) of \( \sin(x) \) over the interval \( [0, \pi] \) is \( \frac{2}{\pi} \).