Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 26

$$s= \int_{0}^{1}\sqrt{1+4u^2}du$$

Since $$g(x)=\sqrt{x}\to g'(x)=\frac{1}{2\sqrt{x}}$$ Then \begin{align*} s&=\int_{0}^{1}\sqrt{1+g'^2(x)}dx\\ &= \int_{0}^{1}\sqrt{1+\frac{1}{4x^2}}dx\\ \end{align*} Let $u=\sqrt{x}$; then \begin{align*} s&=\int_{0}^{1}\sqrt{1+\frac{1}{4x^2}}dx\\ &= \int_{0}^{1}\sqrt{1+\frac{1}{4u^2}}2udu\\ &= \int_{0}^{1}\sqrt{1+4u^2}du\\ \end{align*} The last integral represents the arc length of $f(u)=u^2$ on $[0,1]$