Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 1

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

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, $$s=\int_2^{6}\sqrt{1+(y')^2}dx=\int_2^{6}\sqrt{1+16x^6}dx.$$