What is the average value of a function #f(x) = 1/x^2# on the interval [1,3]?

Question

Samantha Adkison · Accepted Answer

#1/3#

The average value of the function #f(x)# on the interval #[a,b]# is

#"average value"=1/(b-a)int_a^bf(x)dx#

Thus, the average value here is

#1/(3-1)int_1^3 1/x^2dx=1/2int_1^3x^-2dx#

To integrate this, use the rule

#intx^ndx=x^(n+1)/(n+1)+C#

Hence we see that

#intx^-2=x^(-2+1)/(-2+1)=x^-1/(-1)=-1/x#

So, when we evaluate #1/2int_1^3x^-2dx#, we get

#1/2[-1/x]_1^3=1/2[-1/3-(-1/1)]=1/2(-1/3+1)#

#=1/3#

Cameron Alexander · Answer

undefined The average value of the function $f(x) = \frac{1}{x^2}$ on the interval $[1,3]$ is given by:

$$ \frac{1}{b - a} \int_a^b f(x) \, dx $$

Substituting $a = 1$ and $b = 3$, and integrating $f(x)$ over the interval $[1,3]$:

$$ \frac{1}{3 - 1} \int_1^3 \frac{1}{x^2} \, dx $$

$$ = \frac{1}{2} \int_1^3 \frac{1}{x^2} \, dx $$

$$ = \frac{1}{2} \left[ -\frac{1}{x} \right]_1^3 $$

$$ = \frac{1}{2} \left( -\frac{1}{3} + \frac{1}{1} \right) $$

$$ = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{3} \right) $$

$$ = \frac{1}{2} \left( \frac{3}{3} - \frac{1}{3} \right) $$

$$ = \frac{1}{2} \left( \frac{2}{3} \right) $$

$$ = \frac{1}{3} $$

So, the average value of $f(x) = \frac{1}{x^2}$ on the interval $[1,3]$ is $\frac{1}{3}$.