Let f1(x)=∫0x f(t)dt,f2(x)=∫0x f1(t)dt and f3(x)=∫0x f2(t)dt if f3(x)=A∫0x f(t)(x−t)2dt then the value of A is

Note that ddxf3(x)=f2(x),ddxf2(x)=f(x) and ddxf1(x)=f(x). Integrating by partsf3(x)=tf2(t)0x−∫0x tf1(t)dt=xf2(x)−∫0x tf1(t)dt=∫0x (x−t)f1(t)dt=−(x−t)22f1(t)0x+12∫0x (x−t)2f(t)dt=12∫0x (x−t)2f(t)dt. Thus  A=12.