Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.4 Exercises - Page 540: 62

$s = 2\sqrt {17} + \frac{1}{2}\ln \left( {\sqrt {17} + 4} \right)$

$$\eqalign{ & y = \frac{1}{2}{x^2}{\text{ on the interval }}\left[ {0,4} \right] \cr & {\text{Differentiate}} \cr & \frac{{dy}}{{dx}} = x \cr & \cr & {\text{Use the arc length formula}} \cr & s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & s = \int_0^4 {\sqrt {1 + {{\left( x \right)}^2}} } dx \cr & s = \int_0^4 {\sqrt {{x^2} + 1} } dx \cr & {\text{Integrate by substitution}}{\text{, from the triangle shown below}}{\text{,}} \cr & x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr & \sqrt {{x^2} + 1} = \sec \theta \cr & {\text{Substituting}} \cr & s = \int {\sec \theta } \left( {{{\sec }^2}\theta } \right)d\theta \cr & s = \int {{{\sec }^3}\theta d\theta } \cr & {\text{Where }}\int {{{\sec }^3}\theta } d\theta = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr & \left( {{\text{see full solution of }}\int {{{\sec }^3}\theta } d\theta {\text{ on the page 519}}} \right) \cr & s = \frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{Write in terms of }}x \cr & s = \frac{1}{2}x\left( {\sqrt {{x^2} + 1} } \right) + \frac{1}{2}\ln \left| {\sqrt {{x^2} + 1} + x} \right| + C \cr & s = \left[ {\frac{1}{2}x\left( {\sqrt {{x^2} + 1} } \right) + \frac{1}{2}\ln \left| {\sqrt {{x^2} + 1} + x} \right|} \right]_0^4 \cr & s = \left[ {\frac{1}{2}\left( 4 \right)\left( {\sqrt {{4^2} + 1} } \right) + \frac{1}{2}\ln \left| {\sqrt {{4^2} + 1} + 4} \right|} \right] - \left[ {\frac{1}{2}\left( 0 \right) + \frac{1}{2}\ln \left| 1 \right|} \right] \cr & {\text{Simplifying}} \cr & s = 2\sqrt {17} + \frac{1}{2}\ln \left( {\sqrt {17} + 4} \right) \cr & s \approx 9.2935 \cr} $$