Answer
$${\text{True}}$$
Work Step by Step
$$\eqalign{
& \int_1^{ + \infty } {{x^{ - 4/3}}} dx \cr
& {\text{Using the definition }}\int_a^{ + \infty } {f\left( x \right)dx} = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)dx} \cr
& \int_1^{ + \infty } {{x^{ - 4/3}}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_1^b {{x^{ - 4/3}}dx} \cr
& {\text{Integrating}} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{{{x^{ - 1/3}}}}{{ - 1/3}}} \right]_1^b \cr
& {\text{ }} = - 3\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{x^{1/3}}}}} \right]_1^b \cr
& {\text{ }} = - 3\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{b^{1/3}}}} - \frac{1}{{{1^{1/3}}}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& {\text{ }} = - 3\left[ {\frac{1}{{{\infty ^{1/3}}}} - \frac{1}{{{1^{1/3}}}}} \right] \cr
& {\text{ }} = - 3\left( {0 - 1} \right) \cr
& {\text{ }} = 3 \cr
& {\text{The statement it is true}} \cr} $$
