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state and prove fundamental theroem of integral calculus.

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

Calculus

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