## Calculus: Early Transcendentals (2nd Edition)

$= \frac{4}{3}\,\left( {{{10}^{\frac{3}{4}}}} \right)$
$\begin{gathered} \int_0^{10} {\frac{{dx}}{{\sqrt[4]{{10 - x}}}}} \hfill \\ \hfill \\ integrate\,\,of\,\,indefinite\,\,integral \hfill \\ \hfill \\ set \hfill \\ 10 - x = u\,\,\,\,then\,\,\,\,\, - dx = du \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\sqrt[4]{{10 - x}}}}\,\, = \,\, - \,\,\int_{}^{} {\frac{1}{{{u^{\frac{1}{4}}}}}} \,du\,} \hfill \\ \hfill \\ {\text{integrate}}\,\,u\sin g\,\,the\,\,power\,\,rule \hfill \\ \hfill \\ = - \frac{4}{3}{u^{\frac{3}{4}}}\, \hfill \\ \hfill \\ = - \frac{4}{3}\,{\left( {10 - x} \right)^{\frac{3}{4}}} \hfill \\ \hfill \\ use\,\,\int_a^b {f\,\left( x \right)} \,dx\,\, = \,\,\mathop {\lim }\limits_{c \to {a^ + }} \int_c^b {f\,\left( x \right)} \,dx \hfill \\ \hfill \\ provided\,\,the\,\,\,limit\,\,exists \hfill \\ \hfill \\ \int_0^{10} {\frac{{dx}}{{\sqrt[4]{{10 - x}}}}} \,\,\, = \,\,\mathop {\lim }\limits_{a \to {{10}^ - }} \,\int_0^a {\frac{{dx}}{{\sqrt[4]{{10 - x}}}}} \hfill \\ \hfill \\ = \,\mathop {\lim }\limits_{a \to {{10}^ - }} \,\,\,\,\left[ { - \frac{4}{3}\,{{\left( {10 - x} \right)}^{\frac{3}{4}}}} \right]_0^a \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = - \frac{4}{3}\mathop {\lim }\limits_{a \to {{10}^ - }} \,\,\,\left[ {\,{{\left( {10 - a} \right)}^{\frac{3}{4}}} - {{10}^{\frac{3}{4}}}} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{4}{3}\,\left( {{{10}^{\frac{3}{4}}}} \right) \hfill \\ \end{gathered}$