The average ordinate of y = sin x over [0, pi | vedclass

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(A) The average value of a continuous function $f(x)$ over the interval $[a, b]$ is given by the formula: $	ext{Average value} = \frac{1}{b-a} \int_a

Vedclass · Accepted Answer

$\frac{2}{\pi}$

Vedclass · Answer

(A) The average value of a continuous function $f(x)$ over the interval $[a, b]$ is given by the formula: $	ext{Average value} = \frac{1}{b-a} \int_a^b f(x) dx$ Here, $f(x) = \sin x$, $a = 0$, and $b = \pi$. Substituting these values into the formula: $	ext{Average ordinate} = \frac{1}{\pi - 0} \int_0^\pi \sin x dx$ $= \frac{1}{\pi} [-\cos x]_0^\pi$ $= \frac{1}{\pi} [-\cos(\pi) - (-\cos(0))]$ $= \frac{1}{\pi} [-(-1) - (-1)]$ $= \frac{1}{\pi} [1 + 1]$ $= \frac{2}{\pi}$ Thus, the correct option is $A$.

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

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