Answer
$$3,050,208$$
Work Step by Step
$$\eqalign{
& \int_0^3 {{m^2}{{\left( {4{m^3} + 2} \right)}^3}} dm \cr
& {\text{use substitution}}{\text{. Let }}u = 4{m^3} + 2,{\text{ so that }}du = 12{m^2}dm,\,\,\,\,\,\,\frac{1}{{12}}du = {m^2}dm \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}m = 3,{\text{ then }}u = 4{\left( 3 \right)^3} + 2 = 110 \cr
& \,\,\,\,\,\,{\text{If }}m = 0,{\text{ then }}u = 4{\left( 0 \right)^3} + 2 = 2 \cr
& {\text{Then}} \cr
& \int_0^3 {{m^2}{{\left( {4{m^3} + 2} \right)}^3}} dm = \int_2^{110} {{u^3}} \left( {\frac{1}{{12}}du} \right) \cr
& = \frac{1}{{12}}\int_2^{110} {{u^3}} du \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \frac{1}{{12}}\left( {\frac{{{u^4}}}{4}} \right)_2^{110} \cr
& = \frac{1}{{48}}\left( {{u^4}} \right)_2^{110} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{{48}}\left( {{{\left( {110} \right)}^4} - {{\left( 2 \right)}^4}} \right) \cr
& = \frac{1}{{48}}\left( {14,610,000 - 16} \right) \cr
& = \frac{{146,409,984}}{{48}} \cr
& = 3,050,208 \cr} $$
