How do you find the average value of f(x)=x/sqrt(x^2+1), 0<=x<=4?

Feb 17, 2017

The average value of $f \left(x\right)$ on 0 ≤ x ≤ 4 is $\frac{1}{4} \left(\sqrt{17} - 1\right)$, which is approximately $0.78$

Explanation:

The average value of a continuous function $f \left(x\right)$ on $\left[a , b\right]$, is given by the formula $A = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$.

$A = \frac{1}{4 - 0} {\int}_{0}^{4} \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$

$A = \frac{1}{4} {\int}_{0}^{4} \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$

This now becomes a trig substitution problem. Let $x = \tan \theta$. Then $\mathrm{dx} = {\sec}^{2} \theta d \theta$. Adjust the bounds of integration accordingly.

$A = \frac{1}{4} {\int}_{0}^{\tan 4} \tan \frac{\theta}{\sqrt{{\tan}^{2} \theta + 1}} \cdot {\sec}^{2} \theta d \theta$

Use the pythagorean identity ${\tan}^{2} \alpha + 1 = {\sec}^{2} \alpha$.

$A = \frac{1}{4} {\int}_{0}^{\tan 4} \tan \frac{\theta}{\sqrt{{\sec}^{2} \theta}} \cdot {\sec}^{2} \theta d \theta$

$A = \frac{1}{4} {\int}_{0}^{\tan 4} \tan \frac{\theta}{\sec} \theta \cdot {\sec}^{2} \theta$

$A = \frac{1}{4} {\int}_{0}^{\tan 4} \tan \theta \sec \theta$

This is a known integral.

$A = \frac{1}{4} {\left[\sec \theta\right]}_{0}^{\tan 4}$

We know from our initial substitution that $\frac{x}{1} = \tan \theta$. This means that $\sqrt{{x}^{2} + 1}$ is the hypotenuse. That's to say $\sec \theta = \sqrt{{x}^{2} + 1}$.

$A = \frac{1}{4} {\left[\sqrt{{x}^{2} + 1}\right]}_{0}^{4}$

Evaluate this using the 2nd fundamental theorem of calculus.

$A = \frac{1}{4} \left(\sqrt{{\left(4\right)}^{2} + 1} - \sqrt{{0}^{2} + 1}\right)$

$A = \frac{1}{4} \sqrt{17} - \frac{1}{4}$

$A = \frac{1}{4} \left(\sqrt{17} - 1\right)$

Hopefully this helps!

Feb 17, 2017

Alternative to get $\frac{1}{4} \left(\sqrt{17} - 1\right)$:

Explanation:

Like HSBC244 said, The average value of a continuous function $f \left(x\right)$ on $\left[a , b\right]$, is given by the formula

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

$A = \frac{1}{4 - 0} {\int}_{0}^{4} \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$

$A = \frac{1}{4} {\int}_{0}^{4} \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$

Instead of doing a trig substitution like HSBC244, you can do a u substitution instead. Set $u = {x}^{2} + 1$ which makes $\mathrm{du} = 2 x$. Multiply the integral by $\frac{2}{2}$ to get the 2 sufficient for the u substitution and put the 2 in the denominator on the outside.

$A = \frac{1}{8} {\int}_{1}^{17} \frac{1}{\sqrt{u}} \mathrm{du}$

Note that the limits of integration changed because we were doing a u substitution. 0 became 1 because:

${0}^{2} + 1 = 1$

Similarly plugging in 4 became 17 because:

${4}^{2} + 1 = 17$

Now rewrite it so it's easier to integrate by treating the square root in the denominator:

$A = \frac{1}{8} {\int}_{1}^{17} {u}^{\frac{1}{2}}$

This integral becomes

$A = \frac{1}{8} {\left[2 \sqrt{u}\right]}_{1}^{17}$

Take the 2 out and plug in 17 and 1. As you can see, this will give you the same answer as HSBC244's answer but does not require a trig substitution.