## Calculus: Early Transcendentals (2nd Edition)

$= \frac{8}{{15}}\,\left( {\sqrt {2} + 1 } \right)$
$\begin{gathered} \int_0^1 {\sqrt {1 + \sqrt x } \,\,dx} \hfill \\ \hfill \\ set\,\,x = {t^2}\,\,\, \to \,\,dx = 2tdt \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \int_0^1 {\sqrt {1 + t} } \,\left( {2tdt} \right)\,\,\, = \,\,\,\,2\int_0^1 {\sqrt {1 + t} \,tdt} \hfill \\ \hfill \\ set\,\,1 + t\, = {u^2}\,\,\, \to \,\,\,dt\, = 2udu\,\,\,and\,\,\,t = {u^2} - 1 \hfill \\ \hfill \\ \int_0^1 {\sqrt {1 + \sqrt x } \,\,dx} = 4\int_1^{\sqrt 2 } {u.\,\left( {{u^2} - 1} \right)\,udu} \, = 4\int_1^{\sqrt 2 } {{u^2}\,\left( {{u^2} - 1} \right)du} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = 4\int_1^{\sqrt 2 } {\,\left( {{u^4} - {u^2}} \right)dy} \,\, \hfill \\ \hfill \\ {\text{ use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,} \hfill \\ \hfill \\ integrate\,\,and\,evaluate \hfill \\ \hfill \\ = 4\,\,\left[ {\frac{{{u^5}}}{5} - \frac{{{u^3}}}{3}} \right]_1^{\sqrt 2 } \hfill \\ \hfill \\ \,use\,\,the\,\,ftc \hfill \\ \hfill \\ = 4\,\left( {\frac{{4\sqrt 2 }}{5} - \frac{{2\sqrt 2 }}{3} - \frac{1}{5} + \frac{1}{3}} \right) = 4\,\left( {\frac{{2\sqrt 2 }}{{15}} + \frac{2}{{15}}} \right) \hfill \\ \hfill \\ = \frac{8}{{15}}\,\left( {\sqrt {2} + 1 } \right) \hfill \\ \end{gathered}$