Average area between $x^n$ and a scaled line through the origin

Consider the functions $x^n$ and $ax$ for $x\ge0$. Specifically, consider the average area between the two curves as $a$ modulates between $0$ and $k$ and refer to it as $\hat a$. What value of $k$ satisfies the condition that $\hat a=nx$? Note that $nx$ is simply the line through the origin and $(1,n)$.

Earlier, I posted a similar question for the special case $n=2$ and solving for $\hat a=x^n$. I became interested in solving the problem in the general case, i.e. for $x^3$, $x^{5/4}$, etc. Notice that in this more general problem statement, we simply need to find $k$ such that $\hat a$ lies on the line through $(0,0)$ and $(1,n)$.

Once again, I intend to answer my own question to practice and improve my mathematical writing. I am trying to become better at mathematical writing so viewing other solutions would be greatly beneficial to myself and others. Additionally, if you choose not to read my answer before coming up with your own, this could potentially become quite interesting as the methodology could vary widely.

In case the wording of the problem is confusing, I modeled a picture of $\hat a$ (in $\color{red}{\text{red}}$) for $n=3$ with $k$ along the horizontal axis. As you can see, the solution for $n=3$ is $k=36$. In $\color{blue}{\text{blue}}$, you can see the line which passes through the origin and $(1,n)$.

Find the intersections of $x^n$ and $ax$ to be $x^n-ax=0\iff x(x^{n-1}-a)=0$, and thus our bounds is $[0,\sqrt[n-1]{a}]$. We can then calculate the area for any specific $a$ as $$I_a=\int_0^{\sqrt[n-1]{a}}\left(x^n-ax\right)dx=\frac{{{a}^{1+\frac{2}{n-1}}}-2{{a}^{\frac{1}{n-1}}}{{\left( {{a}^{\frac{1}{n-1}}}\right) }^{n}}+{{a}^{1+\frac{2}{n-1}}}n}{2n+2}$$Next, we must calculate $\hat a$ as $$\hat a=\frac 1 k\int_0^kI_a= \frac{-{{k}^{2+\frac{2}{n-1}}}+2{{k}^{1+\frac{1}{n-1}+\frac{n}{n-1}}}-2{{k}^{1+\frac{1}{n-1}+\frac{n}{n-1}}}n+{{k}^{2+\frac{2}{n-1}}}{{n}^{2}}}{2kn\left( 2n+2\right) }$$ Now, we need to find the intersection of $\hat a(k, n)$ and the line through the origin and $(1,n)$. \begin{align} &\hat a-nk=0 \\ \implies &\frac{-{{k}^{2+\frac{2}{n-1}}}+2{{k}^{1+\frac{1}{n-1}+\frac{n}{n-1}}}-2{{k}^{1+\frac{1}{n-1}+\frac{n}{n-1}}}n+{{k}^{2+\frac{2}{n-1}}}{{n}^{2}}}{2kn\left( 2n+2\right) }-nk=0 \\ \implies &k=\frac{\left( n-1\right){{2}^{n-1}}}{n\sqrt{n+1}{{\left( \frac{n-1}{n\sqrt{n+1}}\right) }^{n}}} \end{align} And thus, the solution of $k$ such that the average area between $x^n$ and $ax$ for $a$ in the interval $[0,k]$ intersects the line through the origin and $(1,n)$ is exactly $k=\frac{\left( n-1\right){{2}^{n-1}}}{n\sqrt{n+1}{{\left( \frac{n-1}{n\sqrt{n+1}}\right) }^{n}}}$. To put that into perspective, here are the solutions for $k$ for $n\in\mathbb [2,3,4,5,\ldots]$ $$\left\{4\sqrt{3},\ \ 36,\ \ \frac{512}{27}5^{3/2},\ \ \frac{5625}4,\ \ \frac{248832}{3125}{{7}^{\frac{5}{2}}},\ \ \frac{60236288}{729},\ \ \ldots\right\}$$
Starting from Will Sherwood's answer, taking logarithms of both sides $$k=\frac{\left( n-1\right){{2}^{n-1}}}{n\sqrt{n+1}{{\left( \frac{n-1}{n\sqrt{n+1}}\right) }^{n}}}=\left(\frac{2n}{n-1} \right)^{n-1}(n+1)^{\frac{n-1}2}$$ $$\log(k)=(n-1) \log \left(\frac{2 n \sqrt{n+1}}{n-1}\right)$$ expanding the rhs for large values of $n$ leads to $$\log(k)=n \left(\frac{\log (n)}{2}+\log (2)\right)-\frac{1}{2} \left(\log(n) -3+2 \log (2)\right)-\frac{5}{4 n}+\frac{1}{4 n^2}+O\left(\frac{1}{n^{5/2}}\right)$$ which seems to be quite accurate.
For example, using $n=7$, as alredy given by Will Sherwood, $$k=\frac{60236288}{729}\approx 82629$$ while the above approximation leads to $$k=21952 e^{65/49}\approx 82714$$
• What is the benefit to using this to calculate $k$ versus the exact form? I guess that it is less computationally expensive or something similar? – Will Sherwood Aug 8 '16 at 17:16
• @WillSherwood. You gave a good and elegant answer for the problem (I upvoted your answer). What I did, just for the OP's curiosity, is just to show that, for large values of $n$, we can derive a simple approximation easy to evaluate. That is all. Cheers. – Claude Leibovici Aug 9 '16 at 3:58