Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 515: 63

$$L = \frac{{2048 + 1763\sqrt {41} }}{{9375}}$$

$$\eqalign{ & y = {x^{5/4}}{\text{ on the interval }}\left[ {0,1} \right] \cr & {\text{The arc length is given by}} \cr & L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr & y = {x^{5/4}},{\text{ }}\frac{{dy}}{{dx}} = \frac{5}{4}{x^{1/4}},{\text{ then}} \cr & L = \int_0^1 {\sqrt {1 + {{\left( {\frac{5}{4}{x^{1/4}}} \right)}^2}} dx} \cr & L = \int_0^1 {\sqrt {1 + {{\left( {\frac{5}{4}} \right)}^2}\sqrt x } dx} \cr & {\text{Taking the Hint }}{u^2} = 1 + {\left( {\frac{5}{4}} \right)^2}\sqrt x \cr & 2udu = {\left( {\frac{5}{4}} \right)^2}\left( {\frac{1}{{2\sqrt x }}} \right)dx \cr & 2udu = \frac{{25}}{{32\sqrt x }}dx \cr & dx = \frac{{64}}{{25}}u\sqrt x du \cr & dx = \frac{{64}}{{25}}u\left( {{u^2} - 1} \right){\left( {\frac{4}{5}} \right)^2}du \cr & dx = \frac{{1024}}{{625}}u\left( {{u^2} - 1} \right)du \cr & {\text{For }}x = 0 \to u = 1 \cr & {\text{For }}x = 1 \to u = \sqrt {41/16} = \sqrt {41} /4 \cr & {\text{Apply the substitution}} \cr & L = \int_1^{\sqrt {41} /4} {\sqrt {{u^2}} \frac{{1024}}{{625}}u\left( {{u^2} - 1} \right)du} \cr & L = \frac{{1024}}{{625}}\int_1^{\sqrt {41} /4} {\left( {{u^4} - {u^2}} \right)du} \cr & {\text{Integrating}} \cr & L = \frac{{1024}}{{625}}\left[ {\frac{1}{5}{u^5} - \frac{1}{3}{u^3}} \right]_1^{\sqrt {41} /4} \cr & L = \frac{{1024}}{{625}}\left[ {\frac{1}{5}{{\left( {\frac{{\sqrt {41} }}{4}} \right)}^5} - \frac{1}{3}{{\left( {\frac{{\sqrt {41} }}{4}} \right)}^3}} \right] - \frac{{1024}}{{625}}\left[ {\frac{1}{5}{{\left( 1 \right)}^5} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] \cr & L = \frac{{1024}}{{625}}\left[ {\frac{1}{5}\left( {\frac{{1681\sqrt {41} }}{{1024}}} \right) - \frac{1}{3}\left( {\frac{{41\sqrt {41} }}{{64}}} \right)} \right] - \frac{{1024}}{{625}}\left[ { - \frac{2}{{15}}} \right] \cr & L = \frac{{1024}}{{625}}\left[ {\frac{{1681\sqrt {41} }}{{5120}} - \frac{{41\sqrt {41} }}{{192}}} \right] - \frac{{1024}}{{625}}\left[ { - \frac{2}{{15}}} \right] \cr & L = \frac{{1024}}{{625}}\left[ {\frac{{1763\sqrt {41} }}{{15360}}} \right] - \frac{{1024}}{{625}}\left[ { - \frac{2}{{15}}} \right] \cr & L = \frac{{1024}}{{625}}\left[ {\frac{{1763\sqrt {41} }}{{15360}} + \frac{2}{{15}}} \right] \cr & L = \frac{{2048 + 1763\sqrt {41} }}{{9375}} \cr} $$