Question -21

let us take the integral first:- $\int _{ 0 }^{ 1 }{ f(1-k+x)\quad dx } \quad \\ let\quad t=1-k+x\quad ->\quad dt=dx\\ now\quad :-\\ \sum _{ k=1 }^{ n }{ \quad \int _{ 1-k }^{ -k }{ \quad f(t)\quad dt } } \\ now\quad just\quad summing\quad it\quad all\quad up:-\\ ->\quad \int _{ 0 }^{ -1 }{ f(t)\quad dt } +\int _{ -1 }^{ -2 }{ f(t)\quad dt } +\int _{ -2 }^{ -3 }{ f(t)\quad dt } +....+\int _{ 1-n }^{ n }{ f(t)\quad dt } \\ now:-\\ let\quad the\quad antiderivative\quad of\quad f(x)\quad be\quad a\quad function\quad F(X)\\ \therefore \quad \{ F(-1)-F(0)\} +\{ F(-2)-F(-1)\} +\{ F(-3)-F(-2)\} +....\\ ..+\{ F(-n)-F(1-n)\} \\ as\quad most\quad of\quad terms\quad are\quad getting\quad cancelled\quad pairwise\\ we\quad are\quad left\quad with:-\\ -F(0)+F(n)\quad ->\quad F(n)-F(0)\\ \therefore \quad the\quad above\quad fn\quad is\quad nothing\quad but\\ \int _{ 0 }^{ n }{ f(x)\quad dx } \\$