How do you find the average value of the function for f(x)=(3x)/sqrt(1-x^2), -1/2<=x<=1/2?

Jun 29, 2017

The average value is $0$.

Explanation:

The average value of a function $f \left(x\right)$ continuous and defined on $\left[a , b\right]$ is given by

$A = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

So our equation will be

$A = \frac{1}{\frac{1}{2} - \left(- \frac{1}{2}\right)} {\int}_{- \frac{1}{2}}^{\frac{1}{2}} \frac{3 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$A = {\int}_{- \frac{1}{2}}^{\frac{1}{2}} \frac{3 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

We can integrate this using the substitution $u = 1 - {x}^{2}$. Then $\mathrm{du} = - 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{- 2 x}$.

$A = {\int}_{\frac{3}{2}}^{\frac{1}{2}} \frac{3 x}{\sqrt{u} \cdot - 2 x} \mathrm{du}$

$A = - \frac{3}{2} {\int}_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{u}} \mathrm{du}$

$A = - \frac{3}{2} {\int}_{- \frac{1}{2}}^{\frac{1}{2}} {u}^{- \frac{1}{2}} \mathrm{du}$

$A = - \frac{3}{2} {\left[2 {u}^{\frac{1}{2}}\right]}_{- \frac{1}{2}}^{\frac{1}{2}}$

But you can't evaluate this just yet. We haven't reverted to the initial variable, and since we didn't change the bounds, we can't evaluate in $u$.

$A = - \frac{3}{2} {\left[2 {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}\right]}_{- \frac{1}{2}}^{\frac{1}{2}}$

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

$A = - \frac{3}{2} \left(0\right)$

$A = 0$

The average value therefore is $0$. This makes complete sense if you look at the graph--the function is symmetric about the y-axis, and therefore if $| a | = | b |$, then the average value will be $0$.

graph{y = (3x)/sqrt(1 - x^2) [-10, 10, -5, 5]}

Hopefully this helps!