#### Answer

$$\int_2^{6}\sqrt{1+16x^6}dx.$$

#### Work Step by Step

To find the arc length $s$ of the curve $y= x^4$ between $x=2$ and $x=6$, we fist calculate $y'=4 x^3$. Then we have the integral,
$$s=\int_2^{6}\sqrt{1+(y')^2}dx=\int_2^{6}\sqrt{1+16x^6}dx.$$