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Question
state and prove fundamental theroem of integral calculus.

Answer
1. If f is integrable on [a,b] and f = g' for some function g, then int of f from b to a is = g(b) g(a)

Pf.  Let P = {t0,...,tn} be any partition of [a,b]. By The Mean Value Thm. There is a point xi in [t(i-1),ti] st. g(ti) g(t(i-1)) = g'(xi)(ti t(i-1))
         = f(xi)(ti t(i-1)).
If
  mi = inf{f(x): t(i-1) <= x <= ti,
  Mi = sup{f(x): t(i-1) <= x <= ti},
then clearly
         mi(ti t(i-1)) <=f(xi)(ti t(i-1)) <= Mi(ti t(i-1)).

Adding theseequations for I = 1 to n we get

sigma (I =1 to n) mi(ti t(i-1)) <= g(b) g(a) <= sigma(1 to n )Mi(ti t(i-1))

so that L(f,P) <= g(b) g(a) <= U(f,P) for every partition P.

So that g(b) g(a) = int(b to a) f

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