To convert a definite integral into a limit of sum over \([0,1]\), use \(\displaystyle \int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac1n\sum_{r=1}^{n}f\left(\frac{r}{n}\right)\).
Step 1: Recall the limit of a sum definition. For a continuous $f$, $\displaystyle\int_0^1 f(x)\,dx=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^{n}f\!\left(\dfrac{r}{n}\right)$. Step 2: Identify the integrand. Here $f(x)=a^k x^k$, treating $a$ and $k$ as constants. Step 3: Plug in the sample points. $f\!\left(\dfrac{r}{n}\right)=a^k\left(\dfrac{r}{n}\right)^k=\dfrac{a^k r^k}{n^k}$. Step 4: Write the Riemann sum. The integral equals $\displaystyle\lim_{n\to\infty}\dfrac1n\sum_{r=1}^{n}\dfrac{a^k r^k}{n^k}=\lim_{n\to\infty}\dfrac{a^k}{n^{k+1}}\sum_{r=1}^{n}r^k$. Step 5: Expand the power sum. $\sum_{r=1}^{n}r^k=1^k+2^k+\cdots+n^k$. Step 6: Match to the option. So the value is $\displaystyle\lim_{n\to\infty}\dfrac{a^k(1^k+2^k+\cdots+n^k)}{n^{k+1}}$. \[ \boxed{\displaystyle\lim_{n\to\infty}\dfrac{a^k(1^k+2^k+\cdots+n^k)}{n^{k+1}}} \]