Question

solvethis problem  t^2*(sin4t)

Answer 1

as t^2 is always positive so
t∈[0,infinity]-----------(1)
and the domain of x in sin function is -π/2 to +π/2 so domain of sin4t will be 
-π/2 ≤ 4t ≤ +π/2-------(A)
-π/8 ≤ t ≤ π/8
so it will be t∈[-π/8,+π/8]---------(2)
so the common will be t ∈[0,π/8]
now put the values 0 and π/8 in the equation
we will get
t=0 so the equation will be equals to zero 
t=π/8 the answer will be π^2/64
or you can simply the values of - and + π/2 in the step (A)