AL Pure Maths - [x]

11 (a) omitted
if x is integer, then f(x+1)=f(x)=0, otherwise
f(x+1)=x+1-[x+1]-1/2=x+1-[x]-1-1/2=f(x)
f has period 1
(b) when 0<=x<=1,f(x)=x-1/2=>P(x)=(1/2)(x^2-x)
(c) Want P(x+1)=P(x). Consider
P(x+1)=∫f(t) dt + ∫f(t) dt [from x to x+1]
=P(x)+∫f(t) dt [from 0 to 1]
=P(x)+∫(x-1/2) dt [from 0 to 1]
=P(x)+(1/2)(x^2-x) | [0,1]
=P(x)
(c) P(x)=P(x- [x])=(1/2)[(x-[x])^2-(x-[x]))
(d) ∫(P(t)+c) dt=0 [from 0 to 1]
∫((1/2)(t^2-t)+c) dt=0 [from 0 to 1]
(1/6)t^3-(1/4)t^2+ct=0 [from 0 to 1]
(1/6)-(1/4)+c=0
c=1/12
(e)
R(x)=(1/2)∫(t^2-t+1/6) dt [t from 0 to x]
R(x+1)
=R(x)+(1/2)∫(t^2-t+1/6) dt [t from x to x+1]
=R(x)+(1/2)(t^3/3-t^2/2+(1/6)t) | [x,x+1]
=R(x)+(1/2)(x^2+x+1/3-x^2/2-x-1/2+(1/6)x+1/6+x^2/2-(1/6)x)
=R(x)
R(x)
=(1/2)∫(t^2-t+1/6) dt [t from 0 to x]
=(1/2)[x^3/3-x^2/2+(1/6)x]
=x^3/6-x^2/4+x/12
That's all !!!

PART C個度唔係幾明點解
∫f(t) dt [from x to x+1] 變左 ∫f(t) dt [from 0 to 1]