Surface area of revolution is given by, #A# = #2piintf[x]sqrt[1+[f'[x]]^2dx#.
So from the question, #A# = #2piint_1^5[4x^3]sqrt[1+[12x^2]^2dx#
Since #d/dx[4x^3]=12x^2#,[ by the power rule, i.e, #d/dx ax^n=nax^[n-1]#.where #a# is a constant].
We have #A = 2piint_1^5[4x^3]sqrt[1+144x^4 dx].........#[1]# #
let #u = 1+144x^4.....#[2]#,# then #[du]/[dx] = 4[144]x^3#, ie, #[du]/[4[144x^3]]=dx#
substituting for #dx# and #u# in.........#[1]#,
#A = 2piint_1^5[4x^3]sqrt[udu]/[4[144x^3]#, cancelling the the terms in #x^3# and simplifying will result in,
#A = pi/[72]int_1^5 sqrt[udu]#, and after integrating , #A=pi/108sqrt[u^3]#
substituting back for #x# from .....#[2]# #A=pi/108sqrt[[1+144x^4]^3# and when evaluated for the bounds of #x=1# to #x = 5# will give the result in the answer. [ hopefully this was helpful.]
