## Calculus: Early Transcendentals (2nd Edition)

$= 3$
$\begin{gathered} \int_{ - 3}^1 {\frac{{dx}}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}} \hfill \\ \hfill \\ integration\,\,of\,\,indefinite\,\,integral \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}} = \int_{}^{} {\,{{\left( {2x + 6} \right)}^{ - \frac{2}{3}}}dx} \hfill \\ \hfill \\ \operatorname{int} egrate\,\,use\,\,the\,\,power\,\,rule \hfill \\ \hfill \\ = \frac{1}{2}\frac{1}{{\,\left( {\frac{1}{3}} \right)}}\,{\left( {2x + 6} \right)^{\frac{1}{3}}} + C \hfill \\ \hfill \\ use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\ \hfill \\ \int_a^b {f\,\left( x \right)} \,dx\,\, = \,\,\mathop {\lim }\limits_{c \to {a^ + }} \int_c^b {f\,\left( x \right)} \,dx \hfill \\ \hfill \\ provided\,\,the\,\,\,limit\,\,exists \hfill \\ \hfill \\ \int_{ - 3}^1 {\frac{{dx}}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}} \,\, = \mathop {\lim }\limits_{c \to - {3^ + }} \int_c^1 {\frac{1}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}\,dx} \hfill \\ then \hfill \\ \hfill \\ = \mathop {\lim }\limits_{c \to - {3^ + }} \,\left( {\frac{3}{2}\,{{\left( {2x + 6} \right)}^{\frac{1}{3}}}} \right)_c^1 \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \frac{3}{2}\,\left( {2 - \mathop {\lim }\limits_{c \to - {3^ + }} \,{{\left( {2c + 6} \right)}^{\frac{1}{3}}}} \right) \hfill \\ \hfill \\ evaluate\,\,the\,\,limit \hfill \\ \hfill \\ = 3 \hfill \\ \end{gathered}$