Differentiation Question 1

Show as the followings:

(1) If -1 < x < 0, then F(x)=∫_[-1~x] f(t)dt=∫_[-1~x] (t^2+1)dt= (x^3+1)/3+(x+1)
so F'(x)= x^2+1 = f(x)
If 0 < x < 1, then F(x)=∫_[-1~0] f(t)dt+∫_[0~x] f(t)dt = 1/3+ 1+ ∫_[0~x] t^3 dt = 4/3 + x^4 /4
so, F'(x) = x^3 = f(x)
so, F'(x)= f(x) for -1 < x < 0 or 0 < x < 1

(2) F(0)=∫_[-1~0] f(t) dt= 1/3 + 1= 4/3
lim(x - > 0+) [ F(x) - F(0)] /(x-0) = lim(x - > 0+) [ x^4 + 4/3 - 4/3 ]/x =0
lim(x - > 0-) [F(x)-F(0)]/(x-0) = lim(x - > 0-) [ x^3 /3 + x]/x = 1
so, lim(x - > 0) [F(x)-F(0)]/(x-0) does not exist, ie. F(x) is not diff. at x=0