Answer
$$m = \frac{5}{4}$$
Work Step by Step
$$\eqalign{
& \rho \left( x \right) = 1 + {x^3},{\text{ }}0 \leqslant x \leqslant 1 \cr
& {\text{The mass of the object is }}m = \int_a^b {\rho \left( x \right)} dx,{\text{ }}\left( {{\text{See page 460}}} \right) \cr
& m = \int_0^1 {\left( {1 + {x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& m = \left[ {x + \frac{1}{4}{x^4}} \right]_0^1 \cr
& m = \left[ {1 + \frac{1}{4}{{\left( 1 \right)}^4}} \right] - \left[ {0 + \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& m = \frac{5}{4} \cr} $$
