Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 22

$$(b-a) \sqrt{1+m^{2}}$$

By the Pythagorean Theorem \begin{array}{l} {s^{2}=(b-a)^{2}+(m(b-b))^{2}} \\ {\left.s=(c b-a)^{2}+m^{2}(b-a)^{2}\right)^{\frac{1}{2}}} \\ {=(b-a)\left(1 - m^{2}\right)^{\frac{1}{2}}} \end{array} Then, by using arc length \begin{aligned} s &=\int_{a}^{b} \sqrt{1+m^{2}} d x \\ &=[x \sqrt{1+m^{2}}]_{a}^{b} \\ &=b \sqrt{1+m^{2}}-a \sqrt{1+m^{2}} \\ &=(b-a) \sqrt{1+m^{2}} \end{aligned}