Question

Find the average value of the function on the given interval.
f(x) = $\frac{sinx}{x^{2} +1}$, [−π, π].

Answer 1

Question:-

Find the average value of the function on the given interval.

$\sf \large f(x) = \frac{sinx}{x^{2} +1}, \sf \large[−π, π].$

Given:-

$\sf \large f(x) = \frac{sinx}{x^{2} +1}, \sf \large[−π, π].$

To Find:-

• The average value of the function on the given interval.

Solution:-

Formula to get average value of f(x) in the interval

[a, b].

$\sf \large = \frac{1}{b - a} \int ^{b}_{a} \: f(x)dx.$

$\sf \large Here, \: \: f(x) = \frac{sin \: x}{x {}^{2} +1 } \: \: \& \: \: f( - x) = \frac{sin \: x}{x {}^{2} + 1 }$

So, f(x) is an odd function as f(x) + f( – x) = 0.

So, integral would be '0' in [ – π, π].

$\sf \large Here, \: \: f(x) = \frac{sin \: x}{x {}^{2} + 1 }$

$\sf \large The \: average \: value = \frac{1}{2\pi} \int ^{\pi} _{ - \pi} \: \frac{sin \: x}{x {}^{2} + 1 } \: dx$

$\sf \large We \: have \: f(x) = \frac{sin \: x}{x {}^{2} + 1} , \: which \: is \: symmetric \: in \: [−π, π].$

$\sf \large So, \int ^{\pi} _{ - \pi} \: \frac{sin \: x}{x {}^{2} + 1} = 0$

Answer:-

∴ The Average value will be 0.

Hope you have satisfied. ⚘