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WBJEE 2023 Math Solved Paper

Question : 54 of 75

Marks: +1, -0

y=\sin x

[0,\pi]

is :

Solution:

The average value of a function

f(x)

over an interval

[a,b]

is given by the formula:

Average

=\frac{1}{b-a}\int\limits_{a}^{b} f(x)\,dx

In this case, we want to find the average value of the function

y=\sin(x)

over the interval

[0,\pi]

. We can use the formula above, with

f(x)=\sin(x), a=0

, and

b=\pi

Average

=\frac{1}{\pi-0}\int\limits_{0}^{\pi}\sin(x)\,dx

The integral of

\sin(x)

from 0 to

\pi

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

So, the average value is :

Average

=\frac{1}{\pi}\cdot 2=\frac{2}{\pi}

So, the correct answer is Option A,

\frac{2}{\pi}

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