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Question

Mathematics Question on Definite Integral

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

A

2π\frac{2}{\pi}

B

3π\frac{3}{\pi}

C

4π\frac{4}{\pi}

D

π\pi

Answer

2π\frac{2}{\pi}

Explanation

Solution

The correct answer is option 2π\frac{2}{\pi}

Given: μ=1b−a∫abf(x)dx\mu=\frac{1}{b-a}\int_{a}^{b} f(x)dx

μy=1π−σ∫0πsin dx=1π[−cos x]0π\mu_y=\frac{1}{\pi-\sigma}\int_{0}^{\pi}sin\, dx=\frac{1}{\pi}[-cos\,x]^\pi_0

⇒μy=(1+1)π=2π\Rightarrow \mu_y=\frac{(1+1)}{\pi}=\frac{2}{\pi}