Question

How do you find the average value of sinx as x varies between $\left[ 0,\pi \right]$ .

Answer 1

Hint: Now we know that the average value of any function f(x) is given by $\dfrac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx}$ . Now we will substitute the values of f(x), a and b and solve the definite integral using the fundamental theorem of calculus. Hence we have the required average value of the given function.

Complete step by step solution:
We can use definite integral to find the average value of the function.
The average value of the function is the length of the rectangle whose area is exactly the area under the curve f(x).
Now for any function $f\left( x \right)$ in the interval $\left[ a,b \right]$ the average value is given by $\dfrac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx}$ .
Now here we have the given function is sinx.
And the interval is given as $\left[ 0,\pi \right]$ hence we have $a=0$ and $b=\pi $ .
Hence substituting the values in the given formula we have,
$\Rightarrow \dfrac{1}{\pi -0}\int_{0}^{\pi }{\sin xdx}$
Now we know that integration of sinx is - cosx.
Hence we have,
$\Rightarrow \dfrac{1}{\pi }\left[ -\cos x \right]_{0}^{\pi }$
$\Rightarrow \dfrac{1}{\pi }\left[ -\cos \pi -\left( -\cos 0 \right) \right]$
Now we know that $\cos \pi =-1$ and $\cos 0=1$ hence substituting this values we get,
$\begin{align}
  & \Rightarrow \dfrac{1}{\pi }\left[ -\left( -1 \right)-\left( -1 \right) \right] \\
 & \Rightarrow \dfrac{1}{\pi }\left[ +1+1 \right] \\
 & \Rightarrow \dfrac{2}{\pi } \\
\end{align}$
Hence the average of the function sinx in the interval $\left[ 0,\pi \right]$ is $\dfrac{2}{\pi }$.

Note: Now if $y=f\left( x \right)$ is a continuous function on the interval $\left[ a,b \right]$ . Then the mean value theorem tells us that in the interval $\left[ a,b \right]$ we have a point c at which f(x) attains its average value. Hence mean value theorem which states that there exist a point c in the interval such that $f\left( c \right)=\dfrac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx}$ . Also note that on rearranging the terms of $f\left( c \right)=\dfrac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx}$ we get $f\left( c \right)\times \left( b-a \right)=\int_{a}^{b}{f\left( x \right)dx}$ . Now we can see that LHS is a rectangle of height f(c) and width $b-a$ and RHS is an area of $f\left( x \right)$ in the interval $\left[ a,b \right]$ .