Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_{27}^\infty {\frac{2}{{\root 3 \of x }}} dx \cr
& {\text{write the radical as }}{x^{1/3}} \cr
& = \int_{27}^\infty {\frac{2}{{{x^{1/3}}}}} dx \cr
& = \int_{27}^\infty {2{x^{ - 1/3}}} dx \cr
& {\text{solve the improper integral using the definition }}\int_a^\infty {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to \infty } \int_a^b {f\left( x \right)} dx{\text{ }} \cr
& {\text{then}} \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_{27}^b {2{x^{ - 1/3}}} dx{\text{ }} \cr
& {\text{integrate by using the power property }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = 2\mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{x^{2/3}}}}{{2/3}}} \right)_{27}^b \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = 3\mathop {\lim }\limits_{b \to \infty } \left( {{x^{2/3}}} \right)_{27}^b \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = 3\mathop {\lim }\limits_{b \to \infty } \left( {{b^{2/3}} - {{27}^{2/3}}} \right) \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = 3\mathop {\lim }\limits_{b \to \infty } \left( {{b^{2/3}} - 9} \right) \cr
& {\text{evaluate the limit when }}b \to \infty \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = 3\left( {{{\left( \infty \right)}^{2/3}} - 9} \right) \cr
& {\text{Simplify}} \cr
& \int_{27}^\infty {2{x^{ - 1/3}}} dx = \infty \cr
& {\text{Then}} \cr
& {\text{The integral diverges}} \cr} $$
