5.4 Indefinite Integrals and the Net Change Theorem/39

$\int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx$ = $\int _{1}^{64}{\frac {1}{x^{1/2}}}$ + $\int _{1}^{64}{\frac {x^{1/3}}{x^{1/2}}}$

= $\int _{1}^{64}x^{-1/2}+x^{{\frac {1}{3}}-{\frac {1}{2}}}$ = $\int _{1}^{64}x^{-{\frac {1}{2}}}+x^{-1/6}$

Add one to the exponents and divide by the new exponent

= $\int _{1}^{64}{\frac {x^{1/2}}{\frac {1}{2}}}+{\frac {x^{5/6}}{\frac {5}{6}}}$ = $\int _{1}^{64}2x^{\frac {1}{2}}+{\frac {6}{5}}x^{\frac {5}{6}}$

=$2(x)^{\frac {1}{2}}+{\frac {6}{5}}(x)^{\frac {5}{6}}{\bigg |}_{1}^{64}$

= $2(64)^{\frac {1}{2}}+{\frac {6}{5}}(64)^{\frac {5}{6}}-(2(1)^{\frac {1}{2}}+{\frac {6}{5}}(1)^{\frac {5}{6}})$

= $16+38.4-(2+1.2)$

= $54.4-3.2$

= $51.2$

= ${\frac {256}{5}}$