Limit of the bounded area

First, we will find the x-coordinates of intersection of $y=kx+k$ and $y=x^2$ :

$kx+k=x^2$

$x^2-kx-k=0$

$x=\frac{k+\sqrt{k^2+4k}}{2}$ or $x=\frac{k-\sqrt{k^2+4k}}{2}$ .

When x=0: $kx+k=k$ and $x^2=0$ .

As $k$ is a positive number, $y=kx+k$ is above $y=x^2$ between the points of intersection.

$A_k=\int_{\frac{k-\sqrt{k^2+4k}}{2}}^{\frac{k+\sqrt{k^2+4k}}{2}}(kx+k-x^2)dx$ $=-\int_{\frac{k-\sqrt{k^2+4k}}{2}}^{\frac{k+\sqrt{k^2+4k}}{2}}(x^2)dx+\int_{\frac{k-\sqrt{k^2+4k}}{2}}^{\frac{k+\sqrt{k^2+4k}}{2}}(kx)dx+\int_{\frac{k-\sqrt{k^2+4k}}{2}}^{\frac{k+\sqrt{k^2+4k}}{2}}(k)dx$

$=-\frac{1}{3}(\frac{k+\sqrt{k^2+4k}}{2})^3+\frac{1}{3}(\frac{k-\sqrt{k^2+4k}}{2})^3+\frac{1}{2}k(\frac{k+\sqrt{k^2+4k}}{2})^2-\frac{1}{2}k(\frac{k-\sqrt{k^2+4k}}{2})^2+k(\frac{k+\sqrt{k^2+4k}}{2})-k(\frac{k-\sqrt{k^2+4k}}{2})$

$=-\frac{k^2\sqrt{k^2+4k}}{4}-\frac{k^2\sqrt{k^2+4k}}{12}-\frac{k\sqrt{k^2+4k}}{3}+\frac{k^2\sqrt{k^2+4k}}{2}+k\sqrt{k^2+4k}$

$=\frac{k^2\sqrt{k^2+4k}}{6}+\frac{2k\sqrt{k^2+4k}}{3}$

Now, the limit is

$\lim_{k\to\infty}\frac{A_k}{k^3}=\lim_{k\to\infty}\frac{\frac{k^2\sqrt{k^2+4k}}{6}+\frac{2k\sqrt{k^2+4k}}{3}}{k^3}=\lim_{k\to\infty}[(\frac{k^2}{6k^2}+\frac{2k}{3k^2})\times{\frac{\sqrt{k^2+4k}}{k}}]$

$=\lim_{k\to\infty}(\frac{1}{6}-0)\times{\sqrt{\frac{k^2+4k}{k^2}}}=\frac{1}{6}\times{\lim_{k\to\infty}(\sqrt{1+\frac{4}{k}})}=\frac{1}{6}\times1=\boxed{\frac{1}{6}}$ .