1. The problem statement, all variables and given/known data Let f be continuous on an interval I containing 0, and define f1(x) = ∫f(t)dt, f2(x) = ∫f1(t)dt, and in general, fn(x) = ∫fn-1(t)dt for n≥2. Show that fn+1(x) = ∫[(x-t)n/n!]f(t)dt for every n≥0. ALL INTEGRALS DEFINED FROM 0 to x (I can't format :( ) 2. Relevant equations Integration by parts: ∫u dv = uv - ∫vdu. 3. The attempt at a solution I want to do this proof by induction. I am assuming the base step is when n=0 since that is the first possibility. This step is very simple because you can plug 0 into fn+1(x) = ∫[(x-t)n/n!]f(t)dt and get f1(x) = ∫f(t)dt. Now the induction step is where I get messed up. I assume that fn = ∫(x-t)n-1/(n-1)! * f(t)dt. Also, I know that fn= (x-t)n/n! * f(t) based on the fact that fn+1(x) = ∫[(x-t)n/n!]f(t)dt = ∫fn(t)dt. I have tried doing an integral by parts setting u = fn(t) ; du = fn-1(t) ; dv = dt ; and v = t... but I keep ending up with either sequences that don't mean anything or some annoying cyclical stuff where I end up right back where I started.