Chapter 6.3, Problem 52E

To determine

Average value ( k ) of $x^k$ on [0, k ]

Answer to Problem 52E

Average value:

  $k\approx 2.39838$

Explanation of Solution

Given information:

  $k>1$

Such that

  $av\left(f\right)=k$

And

  $a=0,b=k$

And

  $f\left(x\right)=x^k$

Formula for average value of a function:

  $av\left(f\right)=\frac{1}{b-a}{\displaystyle {\int }_a^{b}f\left(x\right)dx}$

Substitute

  $av\left(f\right)=k$

And

  $a=0,b=k$

And

  $f\left(x\right)=x^k$

Simplify and isolate the integral:

  $k=\frac{1}{k-0}{\displaystyle {\int }_0^k{x}^{k}dx}$

That becomes

  $k^2={\displaystyle {\int }_0^k{x}^{k}dx}$

Use power rule to evaluate the integral and substitute in the bounds:

  $k^2={\frac{x^{k+1}}{k+1}|}_0^k=\frac{k^{k+1}}{k+1}-\frac{0^{k+1}}{k+1}=\frac{k^{k+1}}{k+1}$

Since k represents some value of x ,

The equation of x becomes

  $x^2=\frac{x^{x+1}}{x+1}$

Now,

We are required to find the solution to this equation by graphing

  $y_1=x^2$ (in red)

And

  $y_2=\frac{x^{x+1}}{x+1}$ (in blue)

And find the intersection point.

Thus,

The two curves intersect at

  $x\approx 2.39838$

And

  $x\approx 0.47586$

Hence,

  $k\approx 2.39838$

Since

  $k>1$

Then

  $x\approx 0.47586$ is ignored

  

Now,

Check the average value of $f\left(x\right)=x^k$ on the interval [0, k ].

With

  $k=2.39838$

  $av\left(f\right)=\frac{1}{2.39838}{\displaystyle {\int }_0^{2.39838}{x}^{2.39838}dx}\approx 2.39838\approx k$